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DISCRETE MATHEMATICS

Eusebius Doedel

LOGIC

Introduction. First we introduce some basic concepts needed in ourdiscussion of logic. These will be covered in more detail later.

A set is a collection of “objects” (or “elements”).

EXAMPLES :

• the infinite set of all integers : Z ≡ {· · · ,−2,−1, 0, 1, 2, 3, · · ·}.

• the infinite set of all positive integers : Z+ ≡ {1, 2, 3, · · ·}.

• the infinite set R of all real numbers.

• the finite set {T ,F }, where T denotes “True ” and F “False ”.

• the finite set of alphabetic characters : {a, b, c, · · · , z}.

1

A function (or “map”, or “operator”) is a rule that associates to everyelement of a set one element in another set.

EXAMPLES :

• If S1 = {a, b, c} and S2 = {1, 2} then the associations

a 7→ 2, b 7→ 1, c 7→ 2,

define a function f from S1 to S2 . We write

f : S1 −→ S2 .

• Similarly f(n) = n2 defines a function f : Z+ −→ Z

+.

• f(n) = n2 can also be viewed as a function f : Z −→ Z.

2

EXAMPLE :

Let Pn denote the infinite set of all polynomial functions p(x) of degreen or less with integer coefficients.

• The derivative operator D restricted to elements of Pn can beviewed as a function from Pn to Pn−1,

D : Pn −→ Pn−1, D : p 7→ dp

dx.

For example, if p(x) = x3 + 2x + 1, then

D : x3 + 2x + 1 7→ 3x2 + 2 ,

i.e.,D(x3 + 2x + 1) = 3x2 + 2 .

3

EXAMPLE :

• We can also define functions of several variables, e.g.,

f(x, y) = x + y ,

can be viewed as a function “from Z+ cross Z

+ into Z+”.

We writef : Z

+ × Z+ −→ Z

+ .

4

Basic logical operators.

The basic logical operators

∧ (“and” , “conjunction”)

∨ (“or” , “disjunction”)

¬ (“not” , “negation”)

are defined in the tables below :

p ¬pT FF T

p q p ∨ qT T TT F TF T TF F F

p q p ∧ qT T TT F FF T FF F F

5

Let B ≡ {T ,F }. Then we can view ¬, ∨, and ∧, as functions

¬ : B −→ B, ∨ : B × B −→ B , ∧ : B × B −→ B .

We can also view the arithmetic operators −, +, and ×, as functions

− : Z −→ Z, + : Z × Z −→ Z, ∗ : Z × Z −→ Z,

defined by value tables, for example,

x −x· ·-2 2-1 10 01 -12 -2· ·

6

Logical expressions.

A logical expression (or “proposition”) P (p, q, · · ·) is a function

P : B × B × · · · × B −→ B .

For example,

P1(p, q) ≡ p ∨ ¬q and P2(p, q, r) ≡ p ∧ (q ∨ r)

are logical expressions.

HereP1 : B × B −→ B ,

andP2 : B × B × B −→ B .

7

The values of a logical expression can be listed in a truth table .

EXAMPLE :

p q ¬q p ∨ (¬q)T T F TT F T TF T F FF F T T

8

Analogously, arithmetic expressions such as

A1(x, y) ≡ x + (−y) and A2(x, y, z) ≡ x × (y + z)

can be considered as functions

A1 : R × R −→ R, and A2 : R × R × R −→ R ,

or, equivalently,

A1 : R2 −→ R, and A2 : R

3 −→ R .

9

Two propositions are equivalent if they always have the same values.

EXAMPLE :

¬(p ∨ q) is equivalent to ¬p ∧ ¬q ,

(one of de Morgan’s laws), as can be seen in the table below :

p q p ∨ q ¬(p ∨ q) ¬p ¬q ¬p ∧ ¬qT T T F F F FT F T F F T FF T T F T F FF F F T T T T

10

NOTE :

In arithmetic

−(x + y) is equivalent to (−x) + (−y) ,

i.e., we do not have that

−(x + y) is equivalent to (−x) × (−y) .

Thus the analogy between logic and arithmetic is limited.

11

The three basic logical operators ¬ , ∨ , and ∧ , are all we need.

However, it is very convenient to introduce some additional operators,

much like in arithmetic, where we write x3 to denote x × (x × x).

Three such additional operators are

⊕ “exclusive or”→ “conditional” , “if then”↔ “biconditional” , “if and only if” , “iff”

defined as :

p q p ⊕ q p → q p ↔ qT T F T TT F T F FF T T T FF F F T T

12

EXAMPLE :

Suppose two persons, P and Q, are suspected of committing a crime.

• Let P denote the statement by P that

“Q did it, or we did it together”.

• Let Q denote the statement by Q that

“P did it, or if I did it then we did it together”.

• Suppose we know P always tells the truth and Q always lies.

Who committed the crime ?

NOTE : By “did it” we mean “was involved”.

13

Let p denote “P did it” and let q denote “Q did it”.

Then p and q are logical variables.

We are given that the value of the logical expression

q ∨ (p ∧ q) is True ,

and thatp ∨

(

q → (p ∧ q))

is False .

Equivalently we have that the value of the logical expression

¬(

p ∨(

q → (p ∧ q)))

∧(

q ∨ (p ∧ q))

is True .

Our problem is to find for what values of p and q this is the case.

14

As an analogy from arithmetic, consider the problem of finding the valuesof x and y in Z so that

the value of the arithmetic expression x2 + y is 5 ,

and such that

the value of x + y is 3 ,

i.e., we want to find all solutions of the the simultaneous equations

x2 + y = 5 , x + y = 3 .

(How many solutions are there ?)

15

For the “crime problem” we have the truth table

p q ¬(

p ∨(

q → (p ∧q)))

∧(

q ∨ (p ∧ q))

T T F T T T F T TT F F T T F F F FF T T F F F T T FF F F T T F F F F(1) (2) (6) (5) (4) (3) (9) (8) (7)

The order of evaluation has been indicated at the bottom of the table.

The values of the entire expression are in column (9).

We observe that the expression is True only if p = F and q = T .

Therefore Q was involved in the crime, but P was not.

16

EXERCISE :

Consider the logical expression in the preceding “crime” example.

Find a much simpler, equivalent logical expression.

(It must have precisely the same values as listed in column “(9)”.)

17

EXERCISE :

Suppose three persons, P, Q, and R, are suspects in a crime.

• P states that “Q or R, or both, were involved”.

• Q states that “P or R, or both, were involved”.

• R states that “P or Q, but not both, were involved”.

• Suppose P and Q always tell the truth, while R always lies.

Who were involved in the crime ?

NOTE : there may be more than one solution · · ·

18

EXERCISE :

Construct a truth table for the logical expression

(p ∧ (¬(¬p ∨ q))) ∨ (p ∧ q) .

Based on the truth table find a simpler, equivalent logical expression.

19

A contradiction is a logical expression whose value is always False .

For example p ∧ ¬p is a contradiction :

p ¬p p ∧ ¬pT F FF T F

A tautology is a logical expression that is always True .

For example p ∨ ¬p is a tautology :

p ¬p p ∨ ¬pT F TF T T

A logical expression that is neither a tautology nor a contradictionis called a contingency.

20

EXERCISE :

Verify by truth table that the following expressions are tautologies :

(

(p → q) ∧ p)

→ q ,

(

(p → q) ∧ ¬q)

→ ¬p .

21

NOTATION :

We use the symbol “⇒” to indicate that a conditional statement is atautology.

For example, from the preceding exercise we have

(

(p → q) ∧ p)

⇒ q , (”modus ponens”) ,

(

(p → q) ∧ ¬q)

⇒ ¬p , (”modus tollens”) .

22

As another example we show that(

(p → q) ∧ (¬p → r) ∧ (q → r))

→ r

is a tautology.

p q r(

(p → q) ∧ (¬p → r) ∧ (q → r))

→ r

T T T T T F T T T T T

T T F T T F T F F T F

T F T F F F T F T T T

T F F F F F T F T T F

F T T T T T T T T T T

F T F T F T F F F T F

F F T T T T T T T T T

F F F T F T F F T T F

(1) (2) (3) (5) (9) (6) (7) (10) (8) (11) (4)

The last column (11) consist of True values only. Therefore we can write

(

(p → q) ∧ (¬p → r) ∧ (q → r))

⇒ r

QUESTION : Does it matter which of the two ∧’s is evaluated first?23

EXAMPLE :

Here we illustrate another technique that can sometimes be used to showthat a conditional statement is a tautology.

Consider again the logical expression

P (p, q, r) ⇐⇒ (p → q) ∧ (¬p → r) ∧ (q → r) .

We want to show thatP (p, q, r) ⇒ r ,

i.e., thatP (p, q, r) → r always has value True .

NOTE : We need only show that:

We can’t have that P (p, q, r) is True , while r is False .

24

So suppose that P (p, q, r) is True , while r is False .

(We must show that this cannot happen ! )

Thus all three of

(a) p → q (b) ¬p → F and (c) q → F

have value True .

(c) can only have value True if q = F .

(b) can only have value True if ¬p = F , i.e., if p = T .

But then (a) becomes T → F which has value F .

So indeed, not all three, (a), (b), and (c) can have value True .

QED ! (“quod erat demonstrandum”: “which was to be shown”)

25

NOTE :

This was an example of a proof by contradiction.

( We will give more examples later · · · )

26

EXERCISE :

Use a proof “by contradiction” to prove the following:

(

(p → q) ∧ p)

⇒ q ,

(

(p → q) ∧ ¬q)

⇒ ¬p .

( Already done by truth table. )

27

SOLUTIONS :

(p → q) ∧ p ⇒ q

PROOF : Suppose the LHS is True , but the RHS is False .

Thus p → q and p have value True , but q is False .

Since p → q and p are True it follows that q is True .

But this contradicts the assumption that q is False . QED !

(p → q) ∧ ¬q ⇒ ¬p

PROOF : Suppose the LHS is True , but the RHS is False .

Thus p → q and ¬q are True , but ¬p is False .

Thus p → q is True , q is False , and p is True .

Since p → q is True and q is False it follows that p is False .

But this contradicts the assumption that ¬p is False . QED !

EXERCISE :

Also use a ”direct proof” to prove that

(

(p → q) ∧ p)

⇒ q ,

(

(p → q) ∧ ¬q)

⇒ ¬p .

( In a direct proof one assumes that the LHS is True and then one showsthat the RHS must be True also. )

28

SOLUTIONS :

(p → q) ∧ p ⇒ q

PROOF : Suppose the LHS is True .

Thus p → q and p have value True .

Since p → q and p are True it follows that q is True .

Thus the RHS is True . QED !

(p → q) ∧ ¬q ⇒ ¬p

PROOF : Suppose the LHS is True .

Thus p → q and ¬q are True .

Thus p → q is True and and q is False .

It follows that p is False .

Thus the RHS is True . QED !

NOTATION :

From the definition of the ↔ operator we see that logical expressions

P1(p, q, · · ·) and P2(p, q, · · ·) ,

are equivalent if and only if

P1(p, q, · · ·) ↔ P2(p, q, · · ·)is a tautology.

In this case we write

P1(p, q, · · ·) ⇐⇒ P2(p, q, · · ·) .

29

EXAMPLE :

¬(p ∧ q) is equivalent to ¬p ∨ ¬q ,

i.e.,

¬(p ∧ q) ⇐⇒ (¬p ∨ ¬q) ,

as seen in the truth table

p q ¬ (p ∧ q) ↔ (¬p ∨ ¬q)T T F T T F F FT F T F T F T TF T T F T T T FF F T F T T T T

30

The operators

⊕ , → , and ↔ ,

can be expressed in terms of the basic operators

¬ , ∧ , and ∨ ,

as we will now verify :

p q (p ⊕ q) ↔(

(p ∨ q) ∧ ¬ (p ∧ q))

T T F T T F F TT F T T T T T FF T T T T T T FF F F T F F T F

31

p q (p → q) ↔ (q ∨ ¬p)T T T T T FT F F T F FF T T T T TF F T T T T

p q (p ↔ q) ↔(

(q ∨ ¬p) ∧ (p ∨ ¬q))

T T T T T T T FT F F T F F T TF T F T T F F FF F T T T T T T

32

Thus we can write

p ⊕ q ⇐⇒ (p ∨ q) ∧ ¬(p ∧ q) ,

p → q ⇐⇒ q ∨ ¬p ,

p ↔ q ⇐⇒ (q ∨ ¬p) ∧ (p ∨ ¬q) .

It also follows that

p ↔ q ⇐⇒ (p → q) ∧ (q → p) .

33

Basic logical equivalences.

The fundamental logical equivalences (“laws”) are :

p ∨ q ⇐⇒ q ∨ p p ∧ q ⇐⇒ q ∧ p commutative law

p ∨ (q ∧ r) ⇐⇒ p ∧ (q ∨ r) ⇐⇒(p ∨ q) ∧ (p ∨ r) (p ∧ q) ∨ (p ∧ r) distributive law

p ∨ F ⇐⇒ p p ∧ T ⇐⇒ p identity law

p ∨ ¬p ⇐⇒ T p ∧ ¬p ⇐⇒ F complement law

34

Some useful additional laws are :

¬T ⇐⇒ F ¬F ⇐⇒ T negation law

p ∨ p ⇐⇒ p p ∧ p ⇐⇒ p idempotent law

p ∨ T ⇐⇒ T p ∧ F ⇐⇒ F domination law

p ∨ (p ∧ q) ⇐⇒ p p ∧ (p ∨ q) ⇐⇒ p absorption law

NOTE : Remember the absorption laws : they can be very useful !

35

Some more laws are :

(p ∨ q) ∨ r ⇐⇒ p ∨ (q ∨ r) (p ∧ q) ∧ r ⇐⇒ p ∧ (q ∧ r) associative law

¬(p ∨ q) ⇐⇒ ¬p ∧ ¬q ¬(p ∧ q) ⇐⇒ ¬p ∨ ¬q de Morgan

¬(¬p) ⇐⇒ p double negation

p → q ⇐⇒ ¬q → ¬p contrapositive

36

• All laws of logic can in principle be proved using truth tables.

• To illustrate the axiomatic nature of the fundamental laws,we prove an additional law using only the fundamental laws.

• At every step the fundamental law used will be indicated.

• Once proved, additional laws may be used in further proofs.

37

EXAMPLE : Proof of the idempotent law

p ∨ p ⇐⇒ p

p ⇐⇒ p ∨ F identity law

⇐⇒ p ∨ (p ∧ ¬p) complement law

⇐⇒ (p ∨ p) ∧ (p ∨ ¬p) distributive law

⇐⇒ (p ∨ p) ∧ T complement law

⇐⇒ p ∨ p identity law

38

NOTE :

• Proving additional laws, using only the fundamental laws, is not aseasy as one might expect, because we have very few tools available.

• However after proving some of these additional equivalences wehave a more powerful set of laws.

39

Simplification of logical expressions.

It is useful to simplify logical expressions as much as possible.

This is much like in arithmetic where, for example, the expression

x3 + 3x2y + 3xy2 + y3

is equivalent to

(x + y)3 .

40

EXAMPLE : Reconsider the logical expression

¬(

p ∨(

q → (p ∧ q)))

∧(

q ∨ (p ∧ q))

from the “crime example”.

It is equivalent to the much simpler logical expression

¬p ∧ q ,

because it has the same truth table values :

p q ¬p ¬p ∧ qT T F FT F F FF T T TF F T F

41

One way to simplify a logical expression is by using

• the fundamental laws of logic,

• known additional laws,

• the definitions of the additional logical operators.

For the “crime example” this can be done as follows :

42

¬(

p ∨(

q → (p ∧ q)))

∧(

q ∨ (p ∧ q))

⇐⇒ ¬(

p ∨(

q → (p ∧ q)))

∧(

q ∨ (q ∧ p))

commutative law

⇐⇒ ¬(

p ∨(

q → (p ∧ q)))

∧ q absorption law

⇐⇒ ¬(

p ∨(

(p ∧ q) ∨ ¬q))

∧ q equivalence of →

⇐⇒ ¬((

p ∨ (p ∧ q))

∨ ¬q)

∧ q associative law

⇐⇒ ¬(

p ∨ ¬q)

∧ q absorption law

⇐⇒ · · ·

43

¬(

p ∨ ¬q)

∧ q

⇐⇒(

¬p ∧ ¬¬q)

∧ q de Morgan

⇐⇒(

¬p ∧ q)

∧ q double negation

⇐⇒ ¬p ∧(

q ∧ q)

associative law

⇐⇒ ¬p ∧ q idempotent law

44

EXERCISE :

Use logical equivalences to simplify the logical expression

(p ∧ (¬(¬p ∨ q))) ∨ (p ∧ q) .

(This example was already considered before, using a truth table.)

EXERCISE :

Use logical equivalences to verify the following equivalence:

(¬p ∧ q) ∨ (¬q ∧ p) ⇐⇒ (p ∨ q) ∧ ¬(p ∧ q) .

(This was considered before in connection with the ⊕ operator.)

45

SOLUTION :

(p ∧ (¬(¬p ∨ q))) ∨ (p ∧ q)

⇐⇒ (p ∧ (p ∧ ¬q)) ∨ (p ∧ q) de Morgan

⇐⇒ ((p ∧ p) ∧ ¬q) ∨ (p ∧ q) associative law

⇐⇒ (p ∧ ¬q) ∨ (p ∧ q) idempotent law

⇐⇒ p ∧ (¬q ∨ q) distributive law

⇐⇒ p ∧ T complement law

⇐⇒ p identity law

SOLUTION :

(p ∨ q) ∧ ¬(p ∧ q)

⇐⇒ (p ∨ q) ∧ (¬p ∨ ¬q) de Morgan

⇐⇒ ((p ∨ q) ∧ ¬p) ∨ ((p ∨ q) ∧ ¬q) distributive law

⇐⇒ (¬p ∧ (p ∨ q)) ∨ (¬q ∧ (p ∨ q)) commutative law

⇐⇒ ((¬p ∧ p) ∨ (¬p ∧ q)) ∨ ((¬q ∧ p) ∨ (¬q ∧ q)) distributive law

⇐⇒ (F ∨ (¬p ∧ q)) ∨ ((¬q ∧ p) ∨ F ) complement law

⇐⇒ (¬p ∧ q) ∨ (¬q ∧ p) identity law

EXERCISE :

Use logical equivalences to show that the logical expression

(

(p → q) ∧ (¬p → r) ∧ (q → r))

→ r ,

is a tautology, i.e., show that

(

(p → q) ∧ (¬p → r) ∧ (q → r))

⇒ r .

NOTE : Earlier we proved this by truth table and by contradiction.

46

SOLUTION :

(

(p → q) ∧ (¬p → r) ∧ (q → r))

→ r

⇐⇒(

(q ∨ ¬p) ∧ (r ∨ ¬¬p) ∧ (r ∨ ¬q))

→ r equivalence of →

⇐⇒(

(q ∨ ¬p) ∧ (r ∨ p) ∧ (r ∨ ¬q))

→ r double negation

⇐⇒ r ∨ ¬(

(q ∨ ¬p) ∧ (r ∨ p) ∧ (r ∨ ¬q))

equivalence of →

⇐⇒ ¬(

¬r ∧ (q ∨ ¬p) ∧ (r ∨ p) ∧ (r ∨ ¬q))

de Morgan

⇐⇒ ¬(

¬r ∧ ¬r ∧ (q ∨ ¬p) ∧ (r ∨ p) ∧ (r ∨ ¬q))

idempotent

⇐⇒ ¬(

(q ∨ ¬p) ∧ ¬r ∧ (r ∨ p) ∧ ¬r ∧ (r ∨ ¬q))

commutative +

SOLUTION : continued · · ·

¬(

(q ∨ ¬p) ∧ ¬r ∧ (r ∨ p) ∧ ¬r ∧ (r ∨ ¬q))

⇐⇒ ¬(

(q ∨ ¬p) ∧ (¬r ∧ p) ∧ (¬r ∧ ¬q))

distributive +

⇐⇒ ¬(

(q ∨ ¬p) ∧ ¬r ∧ p ∧ ¬q))

idempotent +

⇐⇒ ¬(

(q ∨ ¬p) ∧ ¬r ∧ ¬(q ∨ ¬p))

de Morgan +

⇐⇒ ¬(

¬r ∧ F

)

complement +

⇐⇒ T

Predicate Calculus.

Let S be a set.

A predicate P is a function from S to {T ,F } :

P : S −→ {T ,F } ,

orP : S × S × · · · × S −→ {T ,F } .

orP : S1 × S2 × · · · × Sn −→ {T ,F } .

47

EXAMPLE : Let Z+ denote the set of all positive integers.

DefineP : Z

+ −→ {T ,F }by

P (x) = T if x ∈ Z+ is even ,

P (x) = F if x ∈ Z+ is odd .

We can think of P (x) as the statement

“x is an even integer”,

which can be either True or False .

• What are the values of P (12), P(37), P(-3) ?

48

EXAMPLE :

Let U be a set and S a subset of U .

Let P (x) denote the statement “x ∈ S”, i.e.,

P (x) ⇐⇒ x ∈ S .

ThenP : U −→ {T ,F } .

49

EXAMPLE :

Let P (x, y) denote the statement “x + y = 5”, i.e.,

P (x, y) ⇐⇒ x + y = 5 .

Then we can think of P as a function

P : Z+ × Z

+ −→ {T ,F } .

50

Quantifiers.

For a more compact notation we introduce the quantifiers

∀ , ∃ , and ∃!

DEFINITIONS : Let S be a set and P a predicate, P : S −→ {T ,F } .

Then we define :

∀x ∈ S P (x) means “ P (x) = T for all x ∈ S ”.

∃x ∈ S P (x) means “ there exists an x ∈ S for which P (x) = T ”.

∃!x ∈ S P (x) means “ there is a unique x ∈ S for which P (x) = T ”.

51

If it is clear from the context what S is, then one often simply writes

∀x P (x) , ∃x P (x) , ∃!x P (x) .

If S has a finite number of elements then

∀x P (x) ⇐⇒ P (x1) ∧ P (x2) ∧ · · · ∧ P (xn)

∃x P (x) ⇐⇒ P (x1) ∨ P (x2) ∨ · · · ∨ P (xn)

and

∃!x P (x) ⇐⇒(

P (x1) ∧ ¬P (x2) ∧ ¬P (x3) ∧ ¬ · · · ∧ ¬P (xn))

∨(

¬P (x1) ∧ P (x2) ∧ ¬P (x3) ∧ ¬ · · · ∧ ¬P (xn))

∨ · · ·∨

(

¬P (x1) ∧ ¬P (x2) ∧ ¬P (x3) ∧ ¬ · · · ∧ P (xn))

52

Let S be a set and

P : S × S −→ {T ,F } .

Then∀x, y P (x, y) and ∀x∀y P (x, y)

both mean

∀x(

∀yP (x, y))

.

Thus ∀x, y P (x, y) means

“P (x, y) is True for any choice of x and y ”.

53

Similarly,

∃x, y P (x, y) and ∃x∃y P (x, y)

both mean

∃x(

∃yP (x, y))

.

Thus ∃x, y P (x, y) means

“There exist an x and y for which P (x, y) is True ”.

54

EXAMPLE :

LetP : S × S −→ {T ,F } ,

whereS = { 1 , 2 } .

Then

∀x, y P (x, y) ⇐⇒ P (1, 1) ∧ P (1, 2) ∧ P (2, 1) ∧ P (2, 2) ,

while

∃x, y P (x, y) ⇐⇒ P (1, 1) ∨ P (1, 2) ∨ P (2, 1) ∨ P (2, 2) .

55

EXERCISE : Let

P : Z × Z −→ {T ,F } ,

where P (x, y) denotes

“ x + y = 5 ”.

What are the values of the following propositions ?

∀x, y P (x, y)

∃x, y P (x, y)

∀x ∃!y P (x, y)

∃x ∀y P (x, y)

56

SOLUTION :P : Z × Z −→ {T ,F }

P (x, y) ⇐⇒ x + y = 5 .

∀x, y P (x, y) is False : Take, for example, x = 1, y = 1.

∃x, y P (x, y) is True : Take, for example, x = −1, y = 6.

∀x ∃!y P (x, y) is True : For given x ∈ Z+, y = 5 − x .

∃x ∀y P (x, y) is False : Suppose such an x exists :

Then, in particular, x + 0 = 5 and x + 1 = 5.

Thus x must have both the values 5 and 4, which is impossible.

EXERCISE : Let

P : Z+ × Z

+ × Z+ −→ {T ,F } ,

where P (x, y, z) denotes the statement

“ x2 + y2 = z ”.

What are the values of the following propositions?

∃x, y, z P (x, y, z)

∀x, y, z P (x, y, z)

∀x, y ∃z P (x, y, z)

∀x, z ∃y P (x, y, z)

∀z ∃x, y P (x, y, z)

57

SOLUTION :P : Z

+ × Z+ × Z

+ −→ {T ,F }

P (x, y) ⇐⇒ x2 + y2 = z .

∃x, y, z P (x, y, z) is True : Take, for example, x = 1, y = 1, z = 2.

∀x, y, z P (x, y, z) is False : Take, for example, x = 1, y = 1, z = 3.

∀x, y ∃z P (x, y, z) is True : For given x, y ∈ Z+, take z = x2 + y2 .

∀x, z ∃y P (x, y, z) is False : Take x = 1 and z = 1.

Then there is no y ∈ Z+ so that x2 + y2 = z.

∀z ∃x, y P (x, y, z) is False : Take z = 1.

Then there are no x, y ∈ Z+ so that x2 + y2 = z.

EXAMPLE :

LetP, Q : S −→ {T ,F } ,

whereS = { 1 , 2 } .

Then

∀x(

P (x) ∨ Q(x))

⇐⇒(

P (1) ∨ Q(1))

∧(

P (2) ∨ Q(2))

,

while

∀x, y(

P (x) ∨ Q(y))

⇐⇒

(

P (1)∨Q(1))

∧(

P (1)∨Q(2))

∧(

P (2)∨Q(1))

∧(

P (2)∨Q(2))

.

58

EXERCISE : ( see preceding example · · · ):

Show that

∀x(

P (x) ∨ Q(x))

is not equivalent to

∀x, y(

P (x) ∨ Q(y))

Hint: Take S = {1, 2} , and find predicates P and Q so that one of thepropositions is True and the other one False .

59

SOLUTION : Take

x P (x) Q(x)1 T F2 F T

Then

∀x(

P (x) ∨ Q(x))

⇐⇒(

P (1) ∨ Q(1))

∧(

P (2) ∨ Q(2))

,

⇐⇒ T ∧ T ⇐⇒ T ,

while

∀x, y(

P (x) ∨ Q(y))

⇐⇒(

P (1)∨Q(1))

∧(

P (1)∨Q(2))

∧(

P (2)∨Q(1))

∧(

P (2)∨Q(2))

.

⇐⇒ T ∧ T ∧ F ∧ T ⇐⇒ F .

EXAMPLE : IfP, Q : S −→ {T ,F } ,

whereS = { 1 , 2 } ,

then

∃x(

P (x) ∧ Q(x))

⇐⇒(

P (1) ∧ Q(1))

∨(

P (2) ∧ Q(2))

.

EXAMPLE : If again

P, Q : S −→ {T ,F } ,and

S = { 1 , 2 } ,

then

∀x(

P (x) → Q(x))

⇐⇒(

P (1) → Q(1))

∧(

P (2) → Q(2))

⇐⇒(

¬P (1) ∨ Q(1))

∧(

¬P (2) ∨ Q(2))

.

60

EXAMPLE :

(

∀xP (x))

∨(

∀xQ(x))

6⇐⇒ ∀x(

P (x) ∨ Q(x))

i.e., there are predicates P and Q for which the equivalence not valid.

As a counterexample take

P, Q : Z+ −→ {T ,F } ,

where

P (x) ⇐⇒ “x is even” , and Q(x) ⇐⇒ “x is odd” .

Then the RHS is True but the LHS is False .

61

EXERCISE :

Show that

(

∃xP (x))

∧(

∃xQ(x))

6⇐⇒ ∃x(

P (x) ∧ Q(x))

by giving an example where LHS and RHS have a different logical value.

62

SOLUTION :

(

∃xP (x))

∧(

∃xQ(x))

6⇐⇒ ∃x(

P (x) ∧ Q(x))

PROOF : (of non-equivalence)

Let P (x) denote

“x is even” .

and let Q(x) denote

“x is odd” .

Then the LHS is clearly True and the RHS is clearly False . QED !

Some equivalences (Valid for any propositions P and Q) :

(1) ¬(

∃xP (x))

⇐⇒ ∀x(

¬P (x))

(2)(

∀xP (x))

∧(

∀xQ(x))

⇐⇒ ∀x(

P (x) ∧ Q(x))

(3)(

∀xP (x))

∧(

∀xQ(x))

⇐⇒ ∀x∀y(

P (x) ∧ Q(y))

(4)(

∀xP (x))

∨(

∀xQ(x))

⇐⇒ ∀x∀y(

P (x) ∨ Q(y))

63

EXAMPLE : Proof of Equivalence (1) when the set S is finite .

¬(

∃xP (x))

⇐⇒ ¬(

P (x1) ∨ P (x2) ∨ · · · ∨ P (xn))

⇐⇒ ¬(

P (x1) ∨(

P (x2) ∨ · · · ∨ P (xn)))

⇐⇒ ¬P (x1) ∧ ¬(

P (x2) ∨ · · · ∨ P (xn))

⇐⇒ · · ·

⇐⇒ ¬P (x1) ∧ ¬P (x2) ∧ ¬ · · · ∧ ¬P (xn)

⇐⇒ ∀x¬P (x)

64

EXERCISES :

These equivalences are easily seen to be valid:

• Prove Equivalence 2 :

(

∀xP (x))

∧(

∀xQ(x))

⇐⇒ ∀x(

P (x) ∧ Q(x))

• Prove Equivalence 3 :

(

∀xP (x))

∧(

∀xQ(x))

⇐⇒ ∀x∀y(

P (x) ∧ Q(y))

65

EXAMPLE : Proof of Equivalence (4) .

(

∀xP (x))

∨(

∀xQ(x))

⇐⇒ ∀x∀y(

P (x) ∨ Q(y))

NOTE : A correct proof consist of verifying that

LHS ⇒ RHS and RHS ⇒ LHS

or equivalently

LHS ⇒ RHS and ¬ LHS ⇒ ¬ RHS

66

PROOF :

(i)(

∀xP (x))

∨(

∀xQ(x))

⇒ ∀x∀y(

P (x) ∨ Q(y))

is easily seen to be True by a direct proof (with 2 cases).

(ii) ¬[(

∀xP (x))

∨(

∀xQ(x))]

⇒ ¬∀x∀y(

P (x) ∨ Q(y))

can be rewritten as(

∃x¬P (x))

∧(

∃x¬Q(x))

⇒ ∃x∃y(

¬P (x) ∧ ¬Q(y))

P and Q being arbitrary, we may replace them by ¬P and ¬Q :(

∃xP (x))

∧(

∃xQ(x))

⇒ ∃x∃y(

P (x) ∧ Q(y))

which is easily seen to be True by a direct proof.

QED !

67

EXERCISE :

Prove that the equivalence

(

∀xP (x))

∧(

∃xQ(x))

⇐⇒ ∀x∃y(

P (x) ∧ Q(y))

is valid for all propositions P and Q.

Hint : This proof can be done along the lines of the preceding proof.

68

SOLUTION :

(

∀xP (x))

∧(

∃xQ(x))

⇐⇒ ∀x∃y(

P (x) ∧ Q(y))

PROOF (of the equivalence) :

(i) If the LHS is True then:

P (x) is True for all x, and Q(x0) is True for some x0.

Then, choosing y = x0, we see that the RHS is also True .

(

∀xP (x))

∧(

∃xQ(x))

⇐⇒ ∀x∃y(

P (x) ∧ Q(y))

PROOF (continued · · · ) :

(ii) Next show that ¬ LHS ⇒ ¬ RHS, i.e.,

(

∃x¬P (x))

∨(

∀x¬Q(x))

⇒ ∃x∀y(

¬P (x) ∨ ¬Q(y))

.


∃xP (x))

∨(

∀xQ(x))

⇒ ∃x∀y(

P (x) ∨ Q(y))

. ⋆

It is easily seen that if the LHS of ⋆ is True then so is the RHS.

(Consider the cases: (a) ∃xP (x) is True and (b) ∀xQ(x) is True .)

More equivalences (Valid for any propositions P and Q) :

(5) ¬(

∀xP (x))

⇐⇒ ∃x¬P (x)

(6)(

∃xP (x))

∨(

∃xQ(x))

⇐⇒ ∃x(

P (x) ∨ Q(x))

(7)(

∃xP (x))

∨(

∃xQ(x))

⇐⇒ ∃x∃y(

P (x) ∨ Q(y))

(8)(

∃xP (x))

∧(

∃xQ(x))

⇐⇒ ∃x∃y(

P (x) ∧ Q(y))

69

Equivalences (5)-(8) follow from

• negating equivalences (1)-(4), and

• replacing P by ¬P and Q by ¬Q ..

70

EXAMPLE : Proof of Equivalence (6), using Equivalence (2)

(2)(

∀xP (x))

∧(

∀xQ(x))

⇐⇒ ∀x(

P (x) ∧ Q(x))

¬((

∀xP (x))

∧(

∀xQ(x)))

⇐⇒ ¬∀x(

P (x) ∧ Q(x))

(

∃x¬P (x))

∨(

∃x¬Q(x)))

⇐⇒ ∃x(

¬P (x) ∨ ¬Q(x))

(6)(

∃xP (x))

∨(

∃xQ(x))

⇐⇒ ∃x(

P (x) ∨ Q(x))

71

EXERCISE :

• Prove Equivalence 7 using Equivalence 3.

• Prove Equivalence 8 using Equivalence 4.

72

EXAMPLE (of negating a logical expression) :

¬∃x∀y∀zP (x, y, z) ⇐⇒ ∀x¬(

∀y(

∀zP (x, y, z)))

⇐⇒ ∀x∃y¬(

∀zP (x, y, z))

⇐⇒ ∀x∃y∃z¬P (x, y, z)

73

EXAMPLE (of transforming a logical expression) :

∃x(

P (x) → Q(x))

⇐⇒ ∃x(

¬P (x) ∨ Q(x))

⇐⇒(

∃x¬P (x))

∨(

∃xQ(x))

⋆

⇐⇒(

¬∀xP (x))

∨(

∃xQ(x))

⇐⇒(

∀xP (x))

→(

∃xQ(x))

⋆ This step follows from the earlier Equivalence 6.

74

EXAMPLE : (from Rosen’s book: in detail)

Let

D(x) denote the statement “x is a duck”

P (x) ,, “x is one of my poultry”

O(x) ,, “x is an officer”

W (x) ,, “x is willing to waltz”

75

Statement logic equivalent

Ducks never waltz ¬∃x(

D(x) ∧ W (x))

∀x(

(D(x) → ¬W (x))

Officers always waltz ¬∃x(

(O(x) ∧ ¬W (x))

∀x(

(O(x) → W (x))

All my poultry are ducks ∀x(

(D(x) ∨ ¬P (x))

∀x(

(P (x) → D(x))

My poultry are not officers ∀x¬(

O(x) ∧ P (x))

∀x(

(P (x) → ¬O(x))

QUESTION : Do the first three statements imply the last one?

76

Do the first three statements imply the last one, i.e., is

[(

∀x(D(x) → ¬W (x)))

∧(

∀x(O(x) → W (x)))

∧(

∀x(P (x) → D(x)))]

→(

∀x(P (x) → ¬O(x)))

a tautology, i.e., do we have

[(

∀x(D(x) → ¬W (x)))

∧(

∀x(O(x) → W (x)))

∧(

∀x(P (x) → D(x)))]

⇒(

∀x(P (x) → ¬O(x)))

The answer is YES .

In fact, it is a tautology for any predicates D, O, P , W .

77

[(

∀x(D(x) → ¬W (x)))

∧(

∀x(O(x) → W (x)))

∧(

∀x(P (x) → D(x)))]

(1) (2) (3)

⇒(

∀x(P (x) → ¬O(x)))

(4)

To prove this, we must show that :

if (1) , (2) , and (3) are True then (4) is True .

To show (4) is True , we must show:

If, for arbitrary z, P (z) is True then, using (1,2,3), ¬O(z) is True .

78

[(

∀x(D(x) → ¬W (x)))

∧(

∀x(O(x) → W (x)))

∧(

∀x(P (x) → D(x)))]

(1) (2) (3)

⇒(

∀x(P (x) → ¬O(x)))

(4)

PROOF : Let z be an arbitrary element from our set of objects.

Assume P (z) is True .

We must show that ¬O(z) is True , i.e., O(z) is False .

From (3) it follows D(z) is True .

From (1) follows ¬W (z) is True . i.e., W (z) is False .

From (2), since W (z) is False , it follows O(z) is False . QED !

79

EXERCISE (from Rosen):

Express each of the following using predicates and quantifiers:

(1) All clear explanations are satisfactory.

(2) Some excuses are unsatisfactory.

(3) Some excuses are not clear explanations.

Question : Does (3) follow from (1) and (2) ?

Hint : See preceding example.

80

REVIEW EXERCISES.

Problem 1. By truth table check if the ⊕ operator is associative:

(p ⊕ q) ⊕ r ⇐⇒ p ⊕ (q ⊕ r) .

Problem 2. Use logical equivalences to prove that

p → (q → r) ⇐⇒ (p ∧ q) → r .

Problem 3. By contradiction show that the following is a tautology:

[(p ∨ t) ∧ (p → q) ∧ (q → r) ∧ (r → s) ∧ (s → t)] → t .

Problem 4. Use logical equivalences to simplify

([(p ∧ q) ↔ p] → p) → (p → q) .

81

SOLUTION Problem 2 : Use logical equivalences to prove that

p → (q → r) ⇐⇒ (p ∧ q) → r .

PROOF :

p → (q → r)

⇐⇒ p → (¬q ∨ r) equivalence of →

⇐⇒ ¬p ∨ (¬q ∨ r) equivalence of →

⇐⇒ (¬p ∨ ¬q) ∨ r associative law

⇐⇒ ¬(p ∧ q) ∨ r de Morgan

⇐⇒ (p ∧ q) → r equivalence of →

SOLUTION Problem 3 : The following is a tautology:

[(p ∨ t) ∧ (p → q) ∧ (q → r) ∧ (r → s) ∧ (s → t)] → t .

PROOF : Assume the LHS is True and the RHS is False .

We show this leads to a contradiction:

If the RHS is False then t is False .

Since s → t is True , we must have that s is False .

Since r → s is True , we must have that r is False .

Since q → r is True , we must have that q is False .

Since p → q is True , we must have that p is False .

Thus both p and t are False . But, by assumption, p ∨ t is True .

Contradiction ! QED !

SOLUTION Problem 4 : Simplify

(

[(p ∧ q) ↔ p] → p)

→ (p → q) .

First we will simplify(p ∧ q) ↔ p .

Using the result we then simplify

[(p ∧ q) ↔ p] → p .

Finally we simplify

(

[(p ∧ q) ↔ p] → p)

→ (p → q) .

(p ∧ q) ↔ p

⇐⇒(

(p ∧ q) → p)

∧(

p → (p ∧ q))

equivalence of ↔

⇐⇒(

¬(p ∧ q) ∨ p)

∧(

¬p ∨ (p ∧ q))

equivalence of →

⇐⇒ (¬p ∨ ¬q ∨ p) ∧(

¬p ∨ (p ∧ q))

de Morgan+

⇐⇒ T ∧(

¬p ∨ (p ∧ q))

complement+

⇐⇒ ¬p ∨ (p ∧ q) indentity law

⇐⇒ (¬p ∨ p) ∧ (¬p ∨ q) distributive law

⇐⇒ T ∧ (¬p ∨ q) complement law

⇐⇒ ¬p ∨ q indentity law

Next consider :

[(p ∧ q) ↔ p] → p

⇐⇒ (¬p ∨ q) → p preceding equivalence

⇐⇒ ¬(¬p ∨ q) ∨ p equivalence of →

⇐⇒ (p ∧ ¬q) ∨ p de Morgan

⇐⇒ p ∨ (p ∧ ¬q) commutative law law

⇐⇒ p absorption law

Finally we have :

(

[(p ∧ q) ↔ p] → p)

→ (p → q)

⇐⇒ p → (p → q) preceding equivalence

⇐⇒ p → (¬p ∨ q) equivalence of →

⇐⇒ ¬p ∨ (¬p ∨ q) equivalence of →

⇐⇒ ¬p ∨ q idempotent+

⇐⇒ p → q equivalence of →

Thus we have shown that(

[(p ∧ q) ↔ p] → p)

→ (p → q) ⇐⇒ p → q .

Problem 5. Verify the following “laws of inference”, which are usefulin proofs:

Tautology Name

(p ∧ (p → q)) ⇒ q modus ponens

(¬q ∧ (p → q)) ⇒ ¬p modus tollens

((p → q) ∧ (q → r)) ⇒ (p → r) hypothetical syllogism

((p ∨ q) ∧ ¬p) ⇒ q disjunctive syllogism

((p ∨ q) ∧ (¬p ∨ r)) ⇒ (q ∨ r) resolution

82

Problem 6. Express the following statements in predicate logic:

(a) “There is a unique x for which P (x) is True .”

(b) “There is no greatest integer.”

(c) “x0 is the smallest integer for which P (x) is True .”

(d) “Every person has exactly two parents.”

NOTE :

- Let P (x, y) denote “y is a parent of x ” .

- You may use the predicates x ≤ y , x > y , and x 6= y .

83

SOLUTION Problem 6 :

(a) “There is a unique x for which P (x) is True ” :

∃!xP (x) .

(b) “There is no number which is greater than all other numbers” :

∀x∃y (y ≥ x) .

(c) “x0 is the smallest x for which P (x) is True ” :

P (x0) ∧ ∀x(

P (x) → (x0 ≤ x))

.

(d) “Every person has exactly two parents” : ∀y∃x1, x2 :

(x1 6= x2) ∧ (x1 6= y) ∧ (x2 6= y) ∧ P (x1, y) ∧ P (x2, y)

∧ {∀z : P (z, y) → (z = x1) ∨ (z = x2)}.

Problem 7.

Let P (x, y, z) denote the statement

x2 y = z ,

where the universe of discourse of all three variables is the set Z.

What is the truth value of each of the following?

P (1, 1, 1) ∀y, z∃xP (x, y, z)

P (0, 7, 0) ∃!y, z∀xP (x, y, z)

∀x, y, zP (x, y, z) ∀x, y∃zP (x, y, z)

∃x, y, zP (x, y, z) ∀x∃y, zP (x, y, z)

84


P (x, y, z) ⇐⇒ x2 y = z , where x, y, z ∈ Z .

P (1, 1, 1) True

P (0, 7, 0) True

∀x, y, zP (x, y, z) False

∃x, y, zP (x, y, z) True

∀y, z∃xP (x, y, z) False

∃!y, z∀xP (x, y, z) True

∀x, y∃zP (x, y, z) True

∀x∃y, zP (x, y, z) True

Problem 8.

Let P (x, y, z, n) denote the statement

xn + yn = zn ,

where x, y, z, n ∈ Z+.

What is the truth value of each of the following:

P (1, 1, 2, 1) ∀x, y∃!zP (x, y, z, 1)

P (3, 4, 5, 2) ∀z∃x, yP (x, y, z, 1)

P (7, 24, 25, 2) ∃x, y∀zP (x, y, z, 2)

∃!x, y, zP (x, y, z, 2) ∃x, y, zP (x, y, z, 3)

NOTE : One of the above is very difficult!

85


P (x, y, z, n) ⇐⇒ xn + yn = zn , where x, y, z, n ∈ Z+ .

P (1, 1, 2, 1) True

P (3, 4, 5, 2) True

P (7, 24, 25, 2) True

∃!x, y, zP (x, y, z, 2) False

∀x, y∃!zP (x, y, z, 1) True

∀z∃x, yP (x, y, z, 1) False

∃x, y∀zP (x, y, z, 2) False

∃x, y, zP (x, y, z, 3) False (Fermat/Wiles)

Problem 9.

Give an example that shows that

∀x∃yP (x, y) 6⇐⇒ ∃y∀xP (x, y) .

Problem 10.

Prove that

∀x[P (x) → Q(x)] ⇒ [∀xP (x) → ∀xQ(x)].

Problem 11.

Prove that

∀x∃y(P (x) ∨ Q(y)) ⇐⇒ ∀xP (x) ∨ ∃xQ(x) .

86


An example that shows that

∀x∃yP (x, y) 6⇐⇒ ∃y∀xP (x, y) .

Let P (x, y) be the predicate x = y, where x, y ∈ Z.

Then ∀x∃yP (x, y) is True .

(For any given x choose y = x to get P (x, y) to be True .)

On the other hand, ∃y∀xP (x, y) is False in this example.

(For if such a y existed, say, y = y0, then we would have ∀x : x = y0,

which is clearly False .)


∀x[P (x) → Q(x)] ⇒ [∀xP (x) → ∀xQ(x)] .

PROOF :

We must show that if the LHS is True then the RHS is also True .

So assume the LHS is True , that is, ∀x[P (x) → Q(x)] . (1)

Now assume that on the RHS ∀xP (x) is True . (2)

We must show that ∀xQ(x) is True .

To do this let x0 be arbitrary.

By (2) P (x0) is True .

By (1) Q(x0) is True .

Since x0 was arbitrary, it follows that ∀xQ(x) is True .

ALTERNATE SOLUTION of Problem 10 :

∀x[P (x) → Q(x)] → [∀xP (x) → ∀xQ(x)]

⇐⇒ ∀x[¬P (x) ∨ Q(x)] → [∃x¬P (x) ∨ ∀xQ(x)] equivalence of →

⇐⇒ ∃x[P (x) ∧ ¬Q(x)] ∨ [∃x¬P (x) ∨ ∀xQ(x)] equivalence of →

⇐⇒ ∃x[P (x) ∧ ¬Q(x)] ∨ ∃x¬P (x) ∨ ∀xQ(x) associative law

⇐⇒ ∃x[(

P (x) ∧ ¬Q(x))

∨ ¬P (x)] ∨ ∀xQ(x) Property 7 from Table

⇐⇒ ∃x[¬P (x) ∨ ¬Q(x)] ∨ ∀xQ(x) complement law+

⇐⇒ [∃x¬P (x)] ∨ [∃x¬Q(x)] ∨ ∀xQ(x) Property 7 from Table

⇐⇒ [∃x¬P (x)] ∨ [∃x¬Q(x)] ∨ ¬[∃x¬Q(x)] Property 1 from Table

⇐⇒ T complement law+


∀x∃y(P (x) ∨ Q(y)) ⇐⇒ ∀xP (x) ∨ ∃xQ(x) .

PROOF :

(1) If the RHS is True then there are two cases to consider:

(i) If ∀xP (x) is True then any y will make the LHS True .

(ii) If ∃xQ(x) is True , say, Q(x0) is True , then, for any x, picky = x0, to see that the LHS is True .

(2) Next show that if the RHS is False then the LHS is False , i.e.,(

∃x¬P (x))

∧(

∀x¬Q(x))

⇒ ∃x∀y(

¬P (x) ∧ ¬Q(y))

.


∃xP (x))

∧(

∀xQ(x))

⇒ ∃x∀y(

P (x) ∧ Q(y))

,

and this implication is easily seen to be valid.

MATHEMATICAL PROOFS.

• We will illustrate some often used basic proof techniques .

(Some of these techniques we have already seen · · · )

• Several examples will be taken from elementary Number Theory.

87

DEFINITIONS : Let n,m ∈ Z+.

• We call n odd if ∃k ∈ Z : k ≥ 0, n = 2k + 1.

• We call n even if n is not odd. (Then n = 2k for some k ∈ Z+.)

• We say “m divides n”, and write m|n, if n = qm for some q ∈ Z+.

• In this case we call m a divisor of n.

• If m|n then we also say that “n is divisible by m”.

• n (n ≥ 2) is a prime number if its only positive divisors are 1 and n.

• n (n ≥ 2) composite if it is not prime.

• n and m are relatively prime if 1 is their only common divisor.

88

Direct proofs.

Many mathematical statements have the form

“ if P then Q ”i.e.,

P ⇒ Q ,

or, more often,

∀x(

P (x) → Q(x))

,

where P and Q represent specific predicates .

RECALL : a direct proof consists of

• assuming that, for arbitrary x, P (x) is True

• demonstrating that Q(x) is then necessarily True also,

89

PROPOSITION : If n ∈ Z+ is odd then n2 is odd.

REMARK :

This proposition is of the form P ⇒ Q or, more specifically,

∀n ∈ Z+ : P (n) → Q(n) ,

where P and Q are predicates (functions)

P, Q : Z+ −→ {T ,F } ,

namely,

P (n) ⇐⇒ ”n is odd ” , Q(n) ⇐⇒ ”n2 is odd ” .

NOTE : Actually Q(n) ⇐⇒ P (n2) here !

90

If n ∈ Z+ is odd then n2 is odd.

PROOF :

Assume n ∈ Z+ is odd (i.e., assume P (n) = T ).

Then, by definition, n = 2k + 1 for some k ∈ Z, k ≥ 0.

By computation we find

n2 = (2k + 1)2 = 2(2k2 + 2k) + 1 = 2m + 1 ,

where we have defined m ≡ 2k2 + 2k.

Thus, by definition, n2 is odd, i.e., Q(n) = T . QED !

91

PROPOSITION : If n ∈ Z+ is odd then 8 | (n − 1)(n + 1) .

PROOF : If n ∈ Z+ is odd then we can write

n = 2k + 1 for some integer k, k ≥ 0 .

By computation we find

(n − 1)(n + 1) = n2 − 1 = 4k2 + 4k = 4k(k + 1) .

Clearly4 | 4k(k + 1) .

Note, however, that either k is even or k + 1 is even, i.e.,

2 | k or 2 | (k + 1) .

Thus8 | 4k(k + 1) . QED !

92

LEMMA (Needed in the following example · · ·) For x ∈ R , x 6= 0, 1 :

n∑

k=0

xk =1 − xn+1

1 − x, ∀n ≥ 0 , ( Geometric sum ) .

PROOF ( a ”constructive proof” ) :

LetSn =

n∑

k=0

xk .

Then

Sn = 1 + x + x2 + · · · + xn−1 + xn ,

x · Sn = x + x2 + · · · + xn−1 + xn + xn+1 ,

so thatSn − x · Sn = (1 − x) · Sn = 1 − xn+1 ,

from which the formula follows. QED !

93

DEFINITION :

A perfect number is a number that equals the sum of all of its divisors,except the number itself.

EXAMPLES :

6 is perfect :6 = 3 + 2 + 1 ,

and 28 is perfect :

28 = 14 + 7 + 4 + 2 + 1 ,

and so is 496 :

496 = 248 + 124 + 62 + 31 + 16 + 8 + 4 + 2 + 1 .

REMARK : Note that 496 = 24 × 31 .

94

PROPOSITION : Let m ∈ Z+, m > 1.

If 2m − 1 is prime, then n ≡ 2m−1(2m − 1) is perfect,

or, using quantifiers,

∀m ∈ Z+ : 2m − 1 is prime → 2m−1(2m − 1) is perfect .

PROOF : Assume 2m − 1 is prime.

Then the divisors of n = 2m−1(2m − 1) are

1, 2, 22, 23, · · · , 2m−1 ,

and

(2m − 1), 2(2m − 1), 22(2m − 1), · · · , 2m−2(2m − 1), 2m−1(2m − 1) .

The last divisor is equal to n , so we do not include it in the sum.

95

The sum is then

m−1∑

k=0

2k + (2m − 1)m−2∑

k=0

2k =1 − 2m

1 − 2+ (2m − 1)

1 − 2m−1

1 − 2

= (2m − 1) + (2m − 1)(2m−1 − 1)

= (2m − 1) (1 + 2m−1 − 1)

= (2m − 1) 2m−1 = n. QED !

NOTE : We used the formulam

∑

k=0

xk =1 − xm+1

1 − x, (the geometric sum) ,

(valid for x 6= 0, 1).

96

Proving the contrapositive.

It is easy to see (by Truth Table) that

p → q ⇐⇒ ¬q → ¬p .

EXAMPLE :

The statement

“n2 even ⇒ n even”,

proved earlier is equivalent to

“¬(n even) ⇒ ¬(n2 even)”,

i.e., it is equivalent to

n odd ⇒ n2 odd .

97

This equivalence justifies the following :

If we must prove

P ⇒ Q ,

then we may equivalently prove the contrapositive

¬Q ⇒ ¬P .

(Proving the contrapositive is sometimes easier .)

98

PROPOSITION : Let n ∈ Z+, with n ≥ 2.

If the sum of the divisors of n is equal to n + 1 then n is prime.

PROOF : We prove the contrapositive :

If n is not prime then the sum of the divisors can not equal n + 1.

So suppose that n is not prime.

Then n has divisors

1, n, and m, for some m ∈ Z+, m 6= 1, m 6= n ,

and possibly more.

Thus the sum of the divisors is greater than n + 1. QED !

99

Some specific contrapositives.

(p ∧ q) → r ⇐⇒ ¬r → (¬p ∨ ¬q)

(p ∨ q) → r ⇐⇒ ¬r → (¬p ∧ ¬q)

(

∀xP (x))

→ q ⇐⇒ ¬q →(

∃x¬P (x))

(

∃xP (x))

→ q ⇐⇒ ¬q →(

∀x¬P (x))

p → ∀xQ(x) ⇐⇒(

∃x¬Q(x))

→ ¬p

p → ∃xQ(x) ⇐⇒(

∀x¬Q(x))

→ ¬p

100

PROPOSITION : Let n ∈ Z+, with n ≥ 2.

∀a, b ∈ Z+ ( n|a ∨ n|b ∨ n 6 |ab ) ⇒ n is prime .

PROOF : The contrapositive is

If n is not prime then ∃a, b ( n 6 |a ∧ n 6 |b ∧ n|ab ) .

Here the contrapositive is easier to understand and quite easy to prove :

Note that if n is not prime thenn = a b ,

for certain integers a and b, both greater than 1 and less than n.

Clearly n 6 |a , n 6 |b , and n|ab . QED !

101

PROPOSITION : Let n ∈ Z+ . Then

5|n2 ⇒ 5|n ,

PROOF : We prove the contrapositive , i.e.,

5 6 |n ⇒ 5 6 |n2 .

So suppose 5 6 |n.

Then we have the following cases :

n = 5k + 1 ⇒ n2 = 25k2 + 10k + 1 = 5(5k2 + 2k) + 1 ,n = 5k + 2 ⇒ n2 = 25k2 + 20k + 4 = 5(5k2 + 4k) + 4 ,n = 5k + 3 ⇒ n2 = 25k2 + 30k + 9 = 5(5k2 + 6k + 1) + 4 ,n = 5k + 4 ⇒ n2 = 25k2 + 40k + 16 = 5(5k2 + 8k + 3) + 1 ,

for k ∈ Z, k ≥ 0.

This shows that 5 6 |n2 . QED !

102

Proof by contradiction.

To prove a statement P ⇒ Q by contradiction :

• assume P = T and Q = F ,

• show that these assumptions lead to an impossible conclusion

(a “contradiction”).

(We have already seen some proofs by contradiction.)

103

PROPOSITION :

If a prime number is the sum of two prime numbers

then one of these equals 2.

PROOF :

Let p1, p2, and p be prime numbers, with p1 + p2 = p.

Suppose that neither p1 nor p2 is equal to 2.

Then both p1 and p2 must be odd ( and greater than 2 ) .

Hence p = p1 + p2 is even, and greater than 2.

This contradicts that p is prime. QED !

104

PROPOSITION :√

2 is irrational, i.e., if m,n ∈ Z+ then m

n6=

√2.

PROOF : Suppose m,n ∈ Z+ and m

n=

√2.

We may assume m and n are relatively prime (cancel common factors).

Then m =√

2 n ⇒ m2 = 2n2 *⇒ m2 even⇒ m even (proved earlier)⇒ ∃k ∈ Z

+(m = 2k)⇒ 2n2 = m2 = (2k)2 = 4k2 (using * above)⇒ n2 = 2k2

⇒ n2 even⇒ n even

Thus both n and m are even and therefore both are divisible by two.

This contradicts that they are relatively prime. QED !

NOTATION : The “⇒” means that the immediately following state-ment is implied by the preceding statement(s).

105

EXERCISE :

Use a proof by contradiction to show the following:

(p ∨ q) ∧ (p → r) ∧ (q → r) ⇒ r .

(p → q) ∧ (q → r) ⇒ p → r .

Hint : See a similar example earlier in the Lecture Notes.

106

SOLUTION :

(p → q) ∧ (q → r) ⇒ p → r .

PROOF (by contradiction) :

Suppose the LHS is True , but the RHS is False .

Since the RHS is False we have p = T and r = F .

Since the LHS is True we have that both p → q and q → r are True .

Since p = T and p → q is True we have q = T .

Since q = T and q → r is True we have r = T .

But this contradicts that r = F . QED !

PROPOSITION :

If the integers

1, 2, 3, · · · , 10,

are placed around a circle, in any order, then there exist three integersin consecutive locations around the circle that have a sum greater thanor equal to 18.

107

PROOF : (by contradiction)

Suppose any three integers in consecutive locations around the circlehave sum less than 18, that is, less then or equal to 17.

Excluding the number 1, which must be placed somewhere, there remainexactly three groups of three integers in consecutive locations.

The total sum is then less than or equal to 1 + 17 + 17 + 17 = 52.

However, we know that this sum must equal

1 + 2 + 3 + · · · + 10 = 55 .

Hence we have a contradiction. QED !

108

Proof by cases (another example) :

PROPOSITION : Let n ∈ Z+ . Then

6 | n(n + 1)(n + 2) .

PROOF : We always have that 2|n or 2|(n + 1) . (Why ?)

There remain three cases to be considered :

For some k ∈ Z+ , k ≥ 0 :

n = 3k : Then 3|n ,

n = 3k + 1 : Then 3|(n + 2) ,

n = 3k + 2 : Then 3|(n + 1) .

QED !

109

FACT : Any real number x can be uniquely written as

x = n + r ,

wheren ∈ Z and r ∈ R , with 0 ≤ r < 1 .

DEFINITION : We then define the floor of x as

⌊x⌋ ≡ n .

EXAMPLES : ⌊7⌋ = 7 , ⌊−7⌋ = −7 , ⌊π⌋ = 3 , ⌊−π⌋ = −4 .

EXAMPLE : Use a proof by cases to show that

⌊2x⌋ = ⌊x⌋ + ⌊x +1

2⌋ .

110

⌊2x⌋ = ⌊x⌋ + ⌊x + 12⌋ , x = n + r , 0 ≤ r < 1

PROOF :

Case 1 : 0 ≤ r < 12

: Then

0 ≤ 2r < 1 and1

2≤ r +

1

2< 1 .

LHS : ⌊2x⌋ = ⌊2n + 2r⌋ = 2n

RHS : ⌊x⌋ = ⌊n + r⌋ = n

⌊x +1

2⌋ = ⌊n + r +

1

2⌋ = n

so that the identity is satisfied.

111

⌊2x⌋ = ⌊x⌋ + ⌊x + 12⌋ , x = n + r , 0 ≤ r < 1

PROOF : (continued · · · )

Case 2 : 12≤ r < 1 : Then

1 ≤ 2r < 2 and 1 ≤ r +1

2< 1 +

1

2.

LHS : ⌊2x⌋ = ⌊2n + 2r⌋ = 2n + 1

RHS : ⌊x⌋ = ⌊n + r⌋ = n

⌊x +1

2⌋ = ⌊n + r +

1

2⌋ = n + 1

so that the identity is satisfied. QED !

112

Existence proofs.

• Mathematical problems often concern the existence, or non-existence,of certain objects.

• Such problems may arise from the mathematical formulation ofproblems that arise in many scientific areas.

• A proof that establishes the existence of a certain object is calledan existence proof.

113

EXAMPLE : For any positive integer n there exists a sequence

of n consecutive composite integers , i.e.,

∀n ∈ Z+∃m ∈ Z

+ : m + i is composite , i = 1, · · · , n .

For example,

n = 2 : (8,9) are 2 consecutive composite integers (m = 7),

n = 3 : (8,9,10) are 3 three consecutive composite integers (m = 7),

n = 4 : (24,25,26,27) are 4 consecutive composite integers (m = 23),

n = 5 : (32,33,34,35,36) are 5 consecutive composite integers (m = 31).

114

∀n ∈ Z+∃m ∈ Z

+ : m + i is composite , i = 1, · · · , n .

PROOF :

Let m = (n + 1)! + 1.

Then, clearly, the n consecutive integers

(n + 1)! + 1 + 1, (n + 1)! + 1 + 2, · · · , (n + 1)! + 1 + n ,

are composite. (Why ?) QED !

NOTE : This is a constructive existence proof : We demonstratedthe existence of m by showing its value (as a function of n).

115

PROPOSITION : There are infinitely many prime numbers.

The idea of the proof (by contradiction) :

• Assume there is only a finite number of prime numbers.

• Then we’ll show ∃N > 1 ∈ Z+ that is neither prime nor composite .

• But this is impossible !

• Thus there must be infinitely many prime numbers !

REMARK :

We will use the fact than any composite number has a prime divisor .

116

PROOF :

Suppose the total number of primes is finite , say,

p1, p2, · · · , pn .

LetN = p1p2 · · · pn + 1

• Then N cannot be prime. (Why not ?)

• Also, none of the p1, p2, · · · , pn divide N ,

since N divided by pi gives a remainder of 1 , (i = 1, · · · , n) .

• Thus N cannot be composite either ! QED !

117

NOTE :

• This proof is a non-constructive existence proof.

• We proved the existence of an infinite number of primeswithout actually showing them !

118

NOTE :

• There is no general recipe for proving a mathematical statement

and often there is more than one correct proof.

• One generally tries to make a proof as simple as possible,

so that others may understand it more easily.

• Nevertheless, proofs can be very difficult, even for relatively simple

statements such as Fermat’s Last Theorem :

¬∃x, y, z, n ∈ Z+, n ≥ 3 : xn + yn = zn .

119

NOTE :

• There are many mathematical statements that are thought to be

correct, but that have not yet been proved (“open problems ”),

e.g., the “Goldbach Conjecture ” :

“Every even integer greater than 2 is the sum of two prime numbers”.

• Indeed, proving mathematical results is as much an art as it is

a science, requiring creativity as much as clarity of thought.

• An essential first step is always to fully understand the problem.

• Where possible, experimentation with simple examples may help

build intuition and perhaps suggest a possible method of proof.

120

REVIEW EXERCISES.

Problem 1. Use a direct proof to show the following:

(p ∨ q) ∧ (q → r) ∧(

(p ∧ s) → t)

∧(

¬q → (u ∧ s))

∧ ¬r ⇒ t .

(Assuming the left-hand-side is True , you must show that t is True .)

Problem 2. Let n be a positive integer.

Prove the following statement by proving its contrapositive:

”If n3 + 2n + 1 is odd then n is even ”.

121

SOLUTION Problem 1 (Direct proof) :

(p ∨ q) ∧ (q → r) ∧(

(p ∧ s) → t)

∧(

¬q → (u ∧ s))

∧ ¬r ⇒ t .

PROOF :

Assume the LHS is True .

Since ¬r is True we have r is False .

Since q → r is True we have q is False .

Since p ∨ q is True we have p is True .

Since ¬q → (u ∧ s) is True and ¬q is True , u and s are True .

Since (p ∧ s) → t and p and s are True it follows that t is True .

QED !

SOLUTION Problem 2 : Let n be a positive integer.

Prove the following statement by proving its contrapositive:

”If n3 + 2n + 1 is odd then n is even ”.

PROOF :

The contrapositive is

”If n is odd then n3 + 2n + 1 is even ”.

So assume that n is odd, n = 2k + 1, for some k ∈ Z, k ≥ 0 .

Thenn3 + 2n + 1 = (2k + 1)3 + 2(2k + 1) + 1

= 2(4k3 + 6k2 + 5k + 2) ( Check! ) ,

which is even. QED !

Problem 3. Let n be an integer. Show that

3|n2 ⇒ 3|n ,

by proving its contrapositive.

Hint : There are two cases to consider.

Problem 4. Give a direct proof to show the following:

The sum of the squares of any two rational numbers is a rational number.

122


The sum of the squares of any two rational numbers is a rational number.

PROOF : Suppose x and y are rational numbers :

x =p1

q1

and y =p2

q2

.

where p1, q1 and p2, q2 are positive integers.

Thenx2 + y2 =

p21

q21

+p2

2

q22

=p2

1q22 + p2

2q21

q21q

22

.

which is rational.

QED !

Problem 5.

Show that for all positive real x

if x is irrational then√

x is irrational .

by proving the contrapositive .

Problem 6. Use a proof by contradiction to prove the following:

If the integers 1, 2, 3, · · · , 7, are placed around a circle, in any order,then there exist two adjacent integers that have a sum greater than orequal to 9 .

(Can you also give a direct Proof ?)

123


Show that for all positive real x

if x is irrational then√

x is irrational .

PROOF (by contrapositive) : The contrapositive is

if√

x is rational then x is rational .

So suppose√

x is rational :

√x =

p

q, for certain p, q ∈ Z

+ .

Then

x =p2

q2,

which is rational. QED !

SOLUTION Problem 6 (Direct proof) :

If the integers 1, 2, 3, · · · , 7, are placed around a circle, in any order,then there exist two adjacent integers that have a sum greater than orequal to 9 .

PROOF :

The number 7 must be placed somewhere.

The smallest neighbors 7 can possibly have are 1 and 2.

Thus the sum of 7 and the biggest of its neighbors is at least 9

QED !

FACT : Any real number x can be uniquely written as

x = n − r ,where

n ∈ Z and r ∈ R , with 0 ≤ r < 1 .

DEFINITION : We then define the ceiling of x as

⌈x⌉ ≡ n .

EXAMPLES : ⌈7⌉ = 7 , ⌈−7⌉ = −7 , ⌈π⌉ = 4 , ⌈−π⌉ = −3 .

Problem 7.

Is the following equality valid for all positive integers n and m ?

⌊n + m

2⌋ = ⌈n

2⌉ + ⌊m

2⌋ .

If Yes then give a proof. If No then give a counterexample.

124

SET THEORY

Basic definitions.

• Let U be the collection of all objects under consideration.

(U is also called the “universe” of objects under consideration.)

• A set is a collection of objects from U .

125

Let A , B , and C be sets.

x ∈ A ⇐⇒ “x is an element (a member) of A”.

x /∈ A ⇐⇒ ¬(x ∈ A)

A ⊆ B ⇐⇒ ∀x ∈ U : x ∈ A ⇒ x ∈ B subset

A = B ⇐⇒ (A ⊆ B) ∧ (B ⊆ A) equality

The above take values in {T ,F }.

126

The following are set-valued :

A ∪ B ≡ { x ∈ U : (x ∈ A) ∨ (x ∈ B) } union

A ∩ B ≡ { x ∈ U : (x ∈ A) ∧ (x ∈ B) } intersection

A ≡ { x ∈ U : x /∈ A } complement

A − B ≡ { x ∈ U : (x ∈ A) ∧ (x /∈ B) } difference

127

Venn diagram.

This is a useful visual aid for proving set theoretic identities.

EXAMPLE : For the two sides of the identity

(A ∩ B) − C = A ∩ (B − C)

we have the following Venn diagrams :

B B

CC

AA

128

The actual proof of the identity, using the above definitions and the lawsof logic, is as follows :

x ∈(

(A ∩ B) − C)

⇐⇒ x ∈ (A ∩ B) ∧ x /∈ C

⇐⇒ (x ∈ A ∧ x ∈ B) ∧ x /∈ C

⇐⇒ x ∈ A ∧ (x ∈ B ∧ x /∈ C) associative law

⇐⇒ x ∈ A ∧ x ∈ (B − C)

⇐⇒ x ∈ A ∩ (B − C)

129

EXAMPLE :

For the two sides of the identity

A − (B ∪ C) = (A − B) ∩ (A − C)

we have the following Venn diagrams :

B B

CC

AA

130

The actual proof of the identity is as follows :

x ∈(

A − (B ∪ C))

⇐⇒ x ∈ A ∧ x /∈ (B ∪ C)

⇐⇒ x ∈ A ∧ ¬(x ∈ (B ∪ C))

⇐⇒ x ∈ A ∧ ¬(x ∈ B ∨ x ∈ C)

⇐⇒ x ∈ A ∧ x /∈ B ∧ x /∈ C de Morgan

⇐⇒ x ∈ A ∧ x ∈ A ∧ x /∈ B ∧ x /∈ C idempotent law

⇐⇒ x ∈ A ∧ x /∈ B ∧ x ∈ A ∧ x /∈ C commut.+assoc.

⇐⇒ x ∈ (A − B) ∧ x ∈ (A − C)

⇐⇒ x ∈ (A − B) ∩ (A − C)

131

EXAMPLE : A ∩ B = A ∪ B ⇒ A = B

PROOF : ( a direct proof · · · )

Assume (A ∩ B) = (A ∪ B). We must show that A = B.

This is done in two stages :

(i) show A ⊆ B and (ii) show B ⊆ A .

To show (i) :

Let x ∈ A. We must show that x ∈ B.

Since x ∈ A it follows that x ∈ A ∪ B.

Since (A ∩ B) = (A ∪ B) it follows that x ∈ (A ∩ B).

Thus x ∈ B also.

The proof of (ii) proceeds along the same steps.

132

Subsets.

S ⊆ U means S is a subset of a universal set U .

The set of all subsets of U is denoted by 2U or P (U) , the power set .

This name is suggested by the following fact :

If U has n elements then P (U) has 2n elements (sets) .

133

EXAMPLE :

LetU = {1 , 2 , 3} .

Then

P (U) ={

{}, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}

.

We see that P (U) has 23 = 8 elements .

NOTE : The empty set ∅ = {} and U itself are included in P (U).

134

Basic set theoretic identities :

A ∪ B = B ∪ A A ∩ B = B ∩ A commutative laws

A ∪ (B ∩ C) = A ∩ (B ∪ C) =(A ∪ B) ∩ (A ∪ C) (A ∩ B) ∪ (A ∩ C) distributive laws

A ∪ ∅ = A A ∩ U = A identity laws

A ∪ A = U A ∩ A = ∅ complement laws

135

Some additional identities :

U = ∅ ∅ = U

A ∪ A = A A ∩ A = A idempotent laws

A ∪ U = U A ∩ ∅ = ∅ domination laws

A ∪ (A ∩ B) = A A ∩ (A ∪ B) = A absorption laws

136

Some more identities :

(A ∪ B) ∪ C = (A ∩ B) ∩ C =A ∪ (B ∪ C) A ∩ (B ∩ C) associative law

A ∪ B = A ∩ B A ∩ B = A ∪ B de Morgan’s laws

¯A = A involution law

All the preceding identities can be proved using the definitions of settheory and the laws of logic.

Note the close correspondence of these identities to the laws of logic.

137

NOTE : One can also proceed axiomatically by only assuming :

• the existence of a power set P (U) , where U is a universal set ,

• special elements U and ∅ ,

• a unary operator ¯ , and two binary operators ∪ and ∩ ,

• the basic set theoretic identities .

Given this setup one can derive all other set theoretic identities.

Note the close correspondence between the above axiomatic setup andthe axiomatic setup of logic !

138

EXAMPLE :

Prove the idempotent lawA ∪ A = A

using only the basic set theoretic identities :

A = A ∪ ∅ identity law

A ∪ (A ∩ A) complement law

(A ∪ A) ∩ (A ∪ A) distributive law

(A ∪ A) ∩ U complement law

A ∪ A identity law

139

EXAMPLE :

( to illustrate the close relation between Set Theory and Logic · · · )

Using another approach we prove the absorption law :

A ∪ (A ∩ B) = A

Thus we must prove

∀x ∈ U : x ∈ A ∪ (A ∩ B) ⇐⇒ x ∈ A

∀x ∈ U : x ∈ A ∨ x ∈ A ∩ B ⇐⇒ x ∈ A

∀x ∈ U : x ∈ A ∨ (x ∈ A ∧ x ∈ B) ⇐⇒ x ∈ A

140

∀x ∈ U : x ∈ A ∨ (x ∈ A ∧ x ∈ B) ⇐⇒ x ∈ A

Define logical predicates a(x) and b(x) :

a(x) ⇐⇒ x ∈ A , b(x) ⇐⇒ x ∈ B .

Then we must prove

∀x ∈ U : a(x) ∨ (a(x) ∧ b(x)) ⇐⇒ a(x) .

It suffices to prove that, for arbitrary logical variables a and b ,

a ∨ (a ∧ b) ⇐⇒ a .

But this is the absorption law from logic !

141

REVIEW EXERCISES.

For each of the following, determine whether it is valid or invalid.If valid then give a proof. If invalid then give a counterexample.

(1) A ∩ (B ∪ A) = A

(2) A ∪ (B ∩ C) = (A ∪ B) ∩ C

(3) (A ∩ B) ∪ (C ∩ D) = (A ∩ D) ∪ (C ∩ B)

(4) (A ∩ B) ∪ (A ∩ B) = A

(5) A ∪ ((B ∪ C) ∩ A) = A

(6) A − (B ∪ C) = (A − B) ∩ (A − C)

(7) B ∩ C ⊆ A ⇒ (B − A) ∩ (C − A) = ∅

(8) (A ∪ B) − (A ∩ B) = A ⇒ B = ∅

142

SOLUTION Problem 3 : Invalid :

In general, the equality is invalid , i.e.,

(A ∩ B) ∪ (C ∩ D) 6= (A ∩ D) ∪ (C ∩ B) .

Take, for example, sets where

A ∩ D and C ∩ B are empty ,while

A ∩ B and C ∩ D are nonempty .

(Draw a Venn diagram !)

Then (A∩B)∪ (C ∩D) is nonempty, while (A∩D)∪ (C ∩B) is empty.

SOLUTION Problem 7 : Valid :

B ∩ C ⊆ A ⇒ (B − A) ∩ (C − A) = ∅

PROOF : By contradiction:

Suppose the LHS is True , but (B − A) ∩ (C − A) 6= ∅.

Thus there exists a b ∈ (B − A) ∩ (C − A).

Hence b ∈ B, and b ∈ C, but b /∈ A.

In particular, it follows that b ∈ B ∩ C.

Since by assumption B ∩ C ⊆ A it follows that b ∈ A.

But this contradicts the fact that b /∈ A. QED !

SOLUTION Problem 8 : Valid :

(A ∪ B) − (A ∩ B) = A ⇒ B = ∅PROOF : By contradiction: Suppose the LHS is True , but B 6= ∅.Thus there exists a b ∈ B. There are two cases :

(i) b ∈ A : Then b ∈ A ∪ B and b ∈ A ∩ B.

Hence, by definition of set difference, b /∈ (A ∪ B) − (A ∩ B).

Since (A ∪ B) − (A ∩ B) = A it follows that b /∈ A.

But this contradicts the assumption that b ∈ A.

(ii) b /∈ A : Then b ∈ A ∪ B and b /∈ A ∩ B.

Hence, by definition of set difference, b ∈ (A ∪ B) − (A ∩ B).

Since (A ∪ B) − (A ∩ B) = A it follows that b ∈ A.

But this contradicts the assumption that b /∈ A. QED !

FUNCTIONS

DEFINITIONS : Let A and B be sets.

Then f is calleda function from A to B

if to each element of A it associates exactly one element of B .

We writef : A −→ B

and we call A the domain of f and B the codomain of f .

We also define the range of f to be

f(A) ≡ {b ∈ B : b = f(a) for some a ∈ A} .

143

We say that f is :

one-to-one (or injective) iff ∀a1, a2 ∈ A : a1 6= a2 ⇒ f(a1) 6= f(a2)

iff ∀a1, a2 ∈ A : f(a1) = f(a2) ⇒ a1 = a2

onto (or surjective) iff ∀b ∈ B ∃a ∈ A : f(a) = b

iff f(A) = B

bijective iff f is one-to-one and onto

144

EXAMPLE :

LetA = {a, b, c} , B = {1, 2} ,

and let f : A −→ B be defined by

f : a 7→ 1 , f : b 7→ 2 , f : c 7→ 1 .

Then f not one-to-one, but f is onto.

a

b

c

1

2

A B

145

EXAMPLE :

LetA = {a, b} , B = {1, 2, 3} ,

and let f : A −→ B be defined by

f : a 7→ 1 , f : b 7→ 3 .

Then f is one-to-one but not onto.

a 1

b2

3

A B

146

EXAMPLE :

Let A = B = Z+ , and let

f : Z+ −→ Z

+

be defined by

f : n 7→ n(n + 1)

2,

i.e.,

f(n) =n(n + 1)

2.

Then f is one-to-one but not onto.

147

f(n) ≡ n(n + 1)/2

PROOF : (1) f is not onto :

Heref(Z+) = { 1 , 3 , 6 , 10 , 15 , 21 , · · · } ,

so it seems that f is not onto.

To be precise, we show that f(n) can never be equal to 2 :

f(n) = 2 ⇐⇒ n(n + 1)/2 = 2 ⇐⇒ n2 + n − 4 = 0 .

But this quadratic equation has no integer roots.

148

(2) f is one-to-one : Assume that f(n1) = f(n2).

We must show that n1 = n2 :

f(n1) = f(n2) ⇐⇒ n1(n1 + 1)/2 = n2(n2 + 1)/2

⇐⇒ n21 + n1 = n2

2 + n2

⇐⇒ n21 − n2

2 = − (n1 − n2)

⇐⇒ (n1 + n2)(n1 − n2) = − (n1 − n2)

⇐⇒ n1 = n2 or n1 + n2 = − 1

However n1, n2 ∈ Z+. Thus n1 + n2 cannot be negative.

It follows that n1 = n2. QED !

149

EXAMPLE :

Definef : Z × Z −→ Z × Z

or equivalentlyf : Z

2 −→ Z2

byf(m,n) = (m + n , m − n) ,

or equivalently, in matrix multiplication notation

f :

(

mn

)

7→(

1 11 −1

) (

mn

)

.

Then f is one-to-one, but not onto.

150

PROOF :

(i) One-to-one :

Supposef(m1, n1) = f(m2, n2) .

Then(m1 + n1 , m1 − n1) = (m2 + n2 , m2 − n2) ,

i.e.,m1 + n1 = m2 + n2 ,

andm1 − n1 = m2 − n2 .

Add and subtract the equations, and divide by 2 to find

m1 = m2 and n1 = n2 ,

that is,(m1, n1) = (m2, n2) .

Thus f is one-to-one.

151

(ii) Not onto :

Let (s, d) ∈ Z2 be arbitrary. Can we solve

f(m,n) = (s, d) ,

i.e., can we solvem + n = s ,

m − n = d ,

for m,n ∈ Z ?

Add and subtract the two equations, and divide by 2 to get

m =s + d

2and n =

s − d

2.

However, m and n need not be integers, e.g., take s = 1, d = 2.

Thus f is not onto. QED !

152

Given two functions

f : A −→ B and g : B −→ C ,

we can compose them :

(g ◦ f)(a) ≡ g(

f(a))

.

A B C

f g

f(a) g(f(a))a

og f

153

EXAMPLE :

Let A = B = C = Z (all integers), and define f, g : Z −→ Z by

f(n) ≡ n2 + 2n − 1 , g(n) ≡ 2n − 1 .

• Let h1(n) ≡ f(

g(n))

. Then

h1(n) = f(2n − 1) = (2n − 1)2 + 2(2n − 1) − 1 = 4n2 − 2 .

• Let h2(n) ≡ g(

f(n))

. Then

h2(n) = g(n2 + 2n − 1) = 2(n2 + 2n − 1) − 1 = 2n2 + 4n − 3 .

• Let h3(n) ≡ g(

g(n))

. Then

h3(n) = g(2n − 1) = 2(2n − 1) − 1 = 4n − 3 .

154

Inverses. Letf : A −→ B ,

andg : B −→ A .

Then g is called the inverse of f if

∀a ∈ A : g(

f(a))

= a ,

and∀b ∈ B : f

(

g(b))

= b .

If f has an inverse g then we say

f is invertible ,

and we write f−1 for g .

155

EXAMPLE :

Let f : Z −→ Z be defined by f : n 7→ n − 1 , i.e.,

f(n) = n − 1 (”shift operator”) .

• f is one-to-one :

If f(n1) = f(n2) then n1 − 1 = n2 − 1 , i.e., n1 = n2 .

• f is onto :

Given any m ∈ Z , can we find n such that f(n) = m ?

That is, can we find n such that n − 1 = m ?

Easy: n = m + 1 !

156

f(n) = n − 1 , f : Z −→ Z

It follows that f is invertible, with inverse

f−1(m) = m + 1 .

Check :

f(

f−1(m))

= f(m + 1) = (m + 1) − 1 = m ,

f−1(

f(n))

= f−1(n − 1) = (n − 1) + 1 = n .

157

EXAMPLE :

Let f : R −→ R be defined by f : x 7→ 1 − 2x , i.e.,

f(x) = 1 − 2x .

• f is one-to-one :

If f(x1) = f(x2) then 1 − 2x1 = 1 − 2x2 , i.e., x1 = x2 .

• f is onto :

Given any y ∈ R , can we find x such that f(x) = y ?

That is, can we find x such that 1 − 2x = y ?

Easy: x = (1 − y)/2 !

( We actually constructed the inverse in this step : f−1(y) = 1−y2

. )

158

f(x) = 1 − 2x , f : R −→ R

We found that f is invertible, with inverse

f−1(y) =1 − y

2.

Check ( not really necessary · · · ) :

f(

f−1(y))

= f((1 − y)/2) = 1 − 2(

(1 − y)/2)

= y ,

f−1(

f(x))

= f−1(1 − 2x) =1 − (1 − 2x)

2= x .

NOTE : We constructed f−1(y) by solving f(x) = y for x .

159

THEOREM :

f : A −→ B is invertible if and only if f is 1 − 1 and onto .

REMARK :

• It is not difficult to see that this theorem holds for finite sets.

• However, the proof also applies to infinite sets.

160

PROOF :

(1a) First we show that if f is invertible then f is 1 − 1 .

By contradiction: Suppose f is invertible but not 1 − 1.

Since f is not 1 − 1 there exist a1, a2 ∈ A, a1 6= a2, such that

f(a1) = f(a2) ≡ b0 .

Since f is invertible there is a function g : B −→ A such that

g(

(f(a))

) = a, ∀a ∈ A .

In particular

g(

f(a1))

= a1 , and g(

f(a2))

= a2 ,

i.e.,g(b0) = a1 , and g(b0) = a2 .

Thus g is not single-valued (not a function). Contradiction !

161

(1b) Now we show that if f is invertible then f is onto.

By contradiction: Suppose f is invertible but not onto.

Since f is not onto there exists b0 ∈ B such that

f(a) 6= b0, ∀a ∈ A .

Since f is invertible there is a function g : B −→ A such that

f(

g(b))

= b, ∀b ∈ B .

In particular

f(

g(b0))

= b0 , where g(b0) ∈ A .

But this contradicts that f(a) 6= b0, ∀a ∈ A .

162

(2a) Next we show that if f is 1 − 1 and onto then f is invertible.

Define a function g : B −→ A as follows :

Since f is 1 − 1 and onto we have that for any b ∈ B

b = f(a) for some unique a ∈ A .

For each such a ∈ A set

g(b) = a .

Then g : B −→ A , and by construction

f(

g(b))

= f(a) = b .

163

(2b) We still must show that ∀a ∈ A : g(

f(a))

= a .

By contradiction : Suppose g(

f(a0))

6= a0 for some a0 ∈ A .

Define b0 = f(a0) . Then b0 ∈ B and

g(b0) 6= a0 ,

where both g(b0) and a0 lie in A .

Since f is one-to-one it follows that

f(

g(b0))

6= f(a0) ,

i.e.,f(

g(b0))

6= b0 .

But this contradicts the result of (2a) ! QED !

164

EXAMPLE :

• Define f : Z+ −→ Z

+ by

f(n) = n(n − 2)(n − 4) + 4 .

Then f is not one-to-one ; for example, f(2) = f(4) = 4 :

n 1 2 3 4 5 6 7 · · ·f(n) 7 4 1 4 19 52 109 · · ·

Using calculus one can show that f(n) is increasing for n ≥ 3.

Thus f is not onto ; for example,

∀n ∈ Z+ : f(n) 6= 2 .

165

• Now letS = f(Z+) = {1, 4, 7, 19, 52, 109, · · ·} ,

and consider f as a function

f : Z+ −→ S .

Then f is onto, but still not one-to-one, since f(2) = f(4) = 4.

• Finally letD = Z

+ − {2} ,

and consider f as a function

f : D −→ S .

Now f is one-to-one and onto, and hence invertible.

166

EXAMPLE : The floor and ceiling functions.

FACT :

∀x ∈ R ∃ ! n ∈ Z and ∃ ! r ∈ R with 0 ≤ r < 1 such that

x = n + r .

We already defined the floor function, ⌊·⌋ , as

⌊x⌋ = n .

EXAMPLES :

⌊π⌋ = 3 , ⌊e⌋ = 2 , ⌊3⌋ = 3 , ⌊−7/2⌋ = − 4 ,

where e = 2.71828 · · · .

167

FACT :

∀x ∈ R ∃ ! n ∈ Z and ∃ ! r ∈ R with 0 ≤ r < 1 such that

x = n − r .

We already defined the ceiling function, ⌈·⌉ , as

⌈x⌉ = n .

EXAMPLES :

⌈π⌉ = 4 , ⌈e⌉ = 3 , ⌈3⌉ = 3 , ⌈−7/2⌉ = − 3 .

168

We see that

⌊·⌋ : R −→ Z ,

and

⌈·⌉ : R −→ Z .

EXERCISE :

• Is ⌊·⌋ one-to-one? onto? invertible?

• Is ⌈·⌉ one-to-one? onto? invertible?

• Draw the graphs of ⌊·⌋ and ⌈·⌉ .

169

EXAMPLE : Let p , k ∈ Z+ .

Then

⌈ p

k⌉ <

p + k

k.

PROOF :

By definition of the ceiling function we can write

p

k= ⌈ p

k⌉ − r ,

where 0 ≤ r < 1 .

Hence

⌈ p

k⌉ =

p

k+ r <

p

k+ 1 =

p + k

k. QED !

170

EXAMPLE : Show that the linear function

f : R2 −→ R

2

defined byf(x, y) = (x + y , x − y) ,

or, in matrix form,

f :

(

xy

)

7→(

1 11 −1

) (

xy

)

,

is one-to-one and onto.

• One-to-one : Exercise!

Hint : See the earlier example where this function was considered as

f : Z2 −→ Z

2 .

171

• Onto :

f(x, y) = (x + y , x − y) or f :

(

xy

)

7→(

1 11 −1

) (

xy

)

We can construct the inverse by solving

f(x, y) = (s, d) ,

that is, by solving

x + y = s , x − y = d ,

for x, y ∈ R :

x =s + d

2, y =

s − d

2.

Thus the inverse is

g(s, d) = (s + d

2,

s − d

2) or g :

(

sd

)

7→(

12

12

12 −1

2

) (

sd

)

.

172

f(x, y) = (x + y , x − y) , g(s, d) = ( s+d2

, s−d2

)

Check ( not really necessary · · · ) :

f(g(s, d)) = f(s + d

2,

s − d

2)

= (s + d

2+

s − d

2,

s + d

2− s − d

2) = (s, d) ,

and

g(f(x, y)) = g(x + y , x − y)

= ((x + y) + (x − y)

2,

(x + y) − (x − y)

2) = (x, y) .

173

EXAMPLE :

More generally, a function

f : R2 −→ R

2

is linear if it can be written as

f(x, y) = ( a11 x + a12 y , a21 x + a22 y ) ,

or equivalently, as matrix-vector multiplication ,

f :

(

xy

)

7→(

a11 a12

a21 a22

) (

xy

)

,

where aij ∈ R , (i, j = 1, 2) .

174

f : R2 −→ R

2 , f :

(

xy

)

7→(

a11 a12

a21 a22

) (

xy

)

This function is invertible if the determinant

D ≡ a11a22 − a12a21 6= 0 .

In this case the inverse is given by

f−1 :

(

sd

)

7→ 1

D

(

a22 −a12

−a21 a11

)(

sd

)

.

EXERCISE : Check that

f−1(f(x, y)) = (x, y) and f(f−1(s, d)) = (s, d) .

175

Now consider the same linear function

f :

(

nm

)

7→(

a11 a12

a21 a22

) (

nm

)

,

but with aij ∈ Z , (i, j = 1, 2) , and as a function

f : Z2 −→ Z

2 .

Is

f−1 :

(

sd

)

7→ 1

D

(

a22 −a12

−a21 a11

)(

sd

)

.

still the inverse?

176

ANSWER : Not in general !

f is now invertible only if the determinant

• D = a11a22 − a12a21 6= 0,

and

• ∀i, j : D | aij .

In this case f−1 is still given by

f−1 :

(

sd

)

7→ 1

D

(

a22 −a12

−a21 a11

) (

sd

)

.

177

EXERCISE : Show that

f :

(

nm

)

7→(

3 24 3

) (

nm

)

,

is invertible as a function f : Z2 −→ Z

2 .

What is the inverse?

EXERCISE : Show that

f :

(

nm

)

7→(

3 14 3

) (

nm

)

,

is not invertible as a function f : Z2 −→ Z

2 .

178

REVIEW EXERCISES.

Problem 1.

Definef : R −→ R

by

f(x) ≡

1/x if x 6= 0 ,

0 if x = 0 .

• Draw the graph of f .

• Is f one-to-one?

• Is f onto?

• What is f−1 ?

179

Problem 2. Consider a function

f : A −→ B .

For each of the following, can you find a function f that is

(i) one-to-one (ii) onto (iii) one-to-one and onto ?

• A = {1, 2, 3} , B = {1, 2}

• A = {1, 2} , B = {1, 2, 3}

• A = {all even positive integers} , B = {all odd positive integers}

180

Problem 3.

Can you find a function f : Z −→ Z+ that is one-to-one and onto ?

Can you find a function g : Z+ −→ Z that is one-to-one and onto ?

Problem 4. Let Sn be a finite set of n elements.

Show that a function

f : Sn −→ Sn

is one-to-one if and only if it is onto.

181


The function f : Z −→ Z+ defined by

f(x) =

−2n if n < 0 ,

2n + 1 if n ≥ 0 ,

is one-to-one and onto .

The function g : Z+ −→ Z defined by

g(x) =

n−12

if n is odd ,

−n2

if n is even ,

is one-to-one and onto .

NOTE : In fact, g is the inverse of f .

Problem 5. Let f : A −→ B and g : B −→ C be functions.

• Suppose f and g are one-to-one.

Is the composition g ◦ f necessarily one-to-one?

• Suppose the composition g ◦ f is one-to-one.

Are f and g necessarily one-to-one?

• Suppose f and g are onto.

Is the composition g ◦ f necessarily onto?

• Suppose the composition g ◦ f is onto.

Are f and g necessarily onto?

Justify your answers.

182

SOLUTION Problem 5 : f : A −→ B and g : B −→ C

• If f and g are 1 − 1 then g ◦ f is 1 − 1.

PROOF : Suppose g(f(a1)) = g(f(a2)).

Since g is 1 − 1 we have f(a1) = f(a2).

Since f is 1 − 1 we have a1 = a2. QED !

• If g ◦ f is 1 − 1 then f and g are 1 − 1.

FALSE : Let A = {a1} , B = {b1, b2} and C = {c1}.Define f(a1) = b1, g(b1) = c1, g(b2) = c1.

Then g ◦ f and f are 1 − 1, but g is not.

SOLUTION Problem 5 (continued · · · )

• If f and g are onto then g ◦ f is onto.

PROOF : Let c ∈ C.

Since g is onto, there exists b ∈ B such that g(b) = c.

Since f is onto, there exists a ∈ A such that f(a) = b.

Then g(f(a)) = g(b) = c, so g ◦ f is onto. QED !

• If g ◦ f is onto then f and g are onto.

FALSE : Let A = {a1} , B = {b1, b2} and C = {c1},and define f(a1) = b1, g(b1) = c1, g(b2) = c1.

Then g ◦ f and g are onto, but f is not.

Problem 6.

Let P2 denote the set of all polynomials of degree 2 or less ,

i.e., polynomials of the form

p(x) = a x2 + b x + c , a, b, c ∈ R , x ∈ R .

Let P1 denote the set of all polynomials of degree 1 or less ,

i.e., polynomials of the form

p(x) = d x + e , d, e ∈ R , x ∈ R .

Consider the derivative function (or derivative operator )

D : P2 −→ P1 .

183

For example,

D : 3x2 + 7x − 4 7→ 6x + 7 ,

andD : 5x − 2π 7→ 5 .

QUESTIONS :

• Is D indeed a function from P2 to P1 ?

• Is D one-to-one ?

• Is D onto ?

• Does D have an inverse ?

184

Problem 7. If A and B are sets, and if

f : A −→ B ,

then for any subset S of A we define the image of S as

f(S) ≡ {b ∈ B : b = f(a) for some a ∈ S} .

Let S and T be subsets of A . Prove that

• f(S ∪ T ) = f(S) ∪ f(T ) ,

• f(S ∩ T ) ⊆ f(S) ∩ f(T ) .

• Also give an example that shows that in general

f(S ∩ T ) 6= f(S) ∩ f(T ) .

185


• f(S ∪ T ) = f(S) ∪ f(T )

PROOF :

LHS ⊆ RHS :

Let b ∈ f(S ∪ T ). Then ∃a ∈ S ∪ T : f(a) = b.

If a ∈ S then b ∈ f(S) and if a ∈ T then b ∈ f(T ).

In either case b ∈ f(S) ∪ f(T ).

RHS ⊆ LHS :

If b ∈ f(S) ∪ f(T ) then we must consider two cases:

b ∈ f(S), in which case ∃s ∈ S : f(s) = b,orb ∈ f(T ), in which case ∃t ∈ T : f(t) = b.

In either case b ∈ f(S ∪ T ). QED !

SOLUTION Problem 7 (continued · · · )

• f(S ∩ T ) ⊆ f(S) ∩ f(T )

PROOF :

Let b ∈ f(S ∩ T ). Then ∃a ∈ S ∩ T : f(a) = b.

Thus a ∈ S and a ∈ T , so f(a) ∈ f(S) and f(a) ∈ f(T ).

Hence f(a) ∈ f(S) ∩ f(T ). QED !

• In general f(S ∩ T ) 6= f(S) ∩ f(T )

EXAMPLE :

Let A = {s, t}, B = {b}, S = {s}, T = {t}.Define f : A −→ B by f(s) = f(t) = b (so f is not one-to-one).

Then S ∩ T = ∅ (the empty set), and f(S ∩ T ) = f(∅) = ∅.But f(S) = {b} and f(T ) = {b}, so that f(S) ∩ f(T ) = {b}.Thus f(S ∩ T ) ⊆ f(S) ∩ f(T ), but f(S ∩ T ) 6= f(S) ∩ f(T ).

Problem 8. If A and B are sets, and if

f : A −→ B ,

then for any subset S of B we define the pre-image of S as

f−1(S) ≡ {a ∈ A : f(a) ∈ S} .

NOTE : f−1(S) is defined even if f does not have an inverse!

Let S and T be subsets of B . Prove that

• f−1(S ∪ T ) = f−1(S) ∪ f−1(T ) ,

• f−1(S ∩ T ) = f−1(S) ∩ f−1(T ) .

186


• f−1(S ∪ T ) = f−1(S) ∪ f−1(T )

PROOF :

a ∈ f−1(S ∪ T ) ⇐⇒ f(a) ∈ S ∪ T⇐⇒ f(a) ∈ S or f(a) ∈ T⇐⇒ a ∈ f−1(S) or a ∈ f−1(T )⇐⇒ a ∈ f−1(S) ∪ f−1(T ). QED !

• f−1(S ∩ T ) = f−1(S) ∩ f−1(T )

PROOF :

a ∈ f−1(S ∩ T ) ⇐⇒ f(a) ∈ S ∩ T⇐⇒ f(a) ∈ S and f(a) ∈ T⇐⇒ a ∈ f−1(S) and a ∈ f−1(T )⇐⇒ a ∈ f−1(S) ∩ f−1(T ). QED !

THE DIVISION THEOREM :

∀n ∈ Z, ∀d ∈ Z+, ∃! q, r ∈ Z : ( 0 ≤ r < d , n = qd + r ) .

EXAMPLE : If n = 21 and d = 8 then

n = 2 · d + 5 .

Thus, here q = 2 and r = 5 .

• d is called the divisor,

• q is called the quotient; we write q = n div d ,

• r is called the remainder; we write r = n mod d .

187

EXAMPLES :

n = 14 , d = 5 : 14 = 2 · d + 4 , so

14 div 5 = 2 and 14 mod 5 = 4 .

n = −14 , d = 5 : −14 = (−3) · d + 1 , so

−14 div 5 = − 3 and − 14 mod 5 = 1 .

188

Let d ∈ Z+ and n, q, r ∈ Z .

From the definitions of “div” and “mod” it follows that :

PROPERTY 1 : n = (n div d) d + n mod d

Example : 23 = (23 div 7) · 7 + 23 mod 7

PROPERTY 2 : If 0 ≤ r < d then (qd + r) mod d = r

Example : (5 · 7 + 3) mod 7 = 3

PROPERTY 3 : (qd + n mod d) mod d = n mod d

Example : (5 · 7 + 23 mod 7) mod 7 = 23 mod 7

189

PROPERTY 4 : Let a, b ∈ Z and d ∈ Z+ . Then

(ad + b) mod d = b mod d .

PROOF :

By the Division Theorem

b = qd + r, where r = b mod d , with 0 ≤ r < d .

Thus, using Property 2

(ad + b) mod d =(

(a + q)d + r)

mod d = r = b mod d .

QED !

EXAMPLE : (57 · 7 + 13) mod 7 = 13 mod 7 .

190

PROPERTY 5 : Let a ∈ Z and d ∈ Z+. Then

(a mod d) mod d = a mod d .

PROOF : Using Property 3,

(a mod d) mod d = (0 · d + a mod d) mod d = a mod d .

EXAMPLE :

(59 mod 7) mod 7 = 59 mod 7 .

191

PROPERTY 6 :

Let a, b ∈ Z and d ∈ Z+. Then

(a + b) mod d = (a mod d + b mod d) mod d .

EXAMPLE :

(5 + 8) mod 3 = 1 = (5 mod 3 + 8 mod 3) mod 3 .

192

PROOF :


a = qad + ra , where ra = a mod d , with 0 ≤ ra < d ,

b = qbd + rb , where rb = b mod d , with 0 ≤ rb < d .

Thus

(a + b) mod d = (qad + ra + qbd + rb) mod d

=(

(qa + qb)d + ra + rb

)

mod d

= (ra + rb) mod d (using Property 4)

= (a mod d + b mod d) mod d . QED !

193

DEFINITION :

If a, b ∈ Z , d ∈ Z+ , and if

a mod d = b mod d ,

then we also writea ≡ b (mod d) ,

and we say

“a is congruent to b modulo d”.

EXAMPLE :83 ≡ 31 (mod 26) .

Note that 83 − 31 = 52 , which is divisible by 26 , i.e.,

26 | (83 − 31) .

194

PROPOSITION : Let a, b ∈ Z , and d ∈ Z+ . Then

a ≡ b (mod d) if and only if d | (a − b) .

PROOF :

(⇒) First, if a ≡ b (mod d) then, by definition,

a mod d = b mod d .

Hence there exist qa, qb, r ∈ Z , with 0 ≤ r < d , such that

a = qad + r and b = qbd + r (same remainder).

It follows thata − b = (qa − qb) d ,

so that d | (a − b).

195

a ≡ b (mod d) if and only if d | (a − b)

(⇐) Conversely, if d | (a − b) then

a − b = qd ,

i.e. ,a = b + qd ,

for some q ∈ Z .

It follows that

a mod d = (b + qd) mod d = b mod d .

196

PROPOSITION : If a, b ∈ Z , and c, d ∈ Z+ , then

a ≡ b (mod d) ⇒ ac ≡ bc (mod dc) .

PROOF :

a ≡ b (mod d) if and only if d | (a − b) ,

i.e.,a − b = qd , for some q ∈ Z .

Then ac − bc = qdc , so that (dc) | (ac − bc) ,

i.e.,ac ≡ bc (mod dc) .

NOTE : We also have that ac ≡ bc (mod d) .

197

PROPOSITION : Let a, b ∈ Z and d ∈ Z+ . Then

a ≡ b (mod 2d) ⇒ a2 ≡ b2 (mod 4d) .

EXAMPLE : Let a = 13 , b = 7 d = 3 . Then

13 ≡ 7 (mod 2 · 3) ,

i.e.,13 (mod 6) = 7 (mod 6) ,

and

132 ≡ 72 (mod 4 · 3) ,

i.e.,169 (mod 12) = 49 (mod 12) ( Check! ) .

198

a ≡ b (mod 2d) ⇒ a2 ≡ b2 (mod 4d)

PROOF : Suppose a ≡ b (mod 2d) .

Then 2d|(a − b) , i.e., a − b = q2d , for some q ∈ Z .

Thus a and b differ by an even number.

It follows that a and b must be both even or both odd.

Hence a + b must be even, i.e., a + b = 2c for some c ∈ Z .

Then a2 − b2 = (a + b) (a − b) = (2c) (q2d) = cq 4d .

It follows that 4d|(a2 − b2) , i.e., a2 ≡ b2 (mod 4d) . QED !

NOTE : Also a2 ≡ b2 (mod 2d) and a2 ≡ b2 (mod d) .

199

PROPOSITION : If n > 3 then not all of

n , n + 2 , n + 4 ,can be primes.

Idea of the proof : Always one of these three numbers is divisible by 3 .

PROOF. By contradiction : Assume that n > 3 and that

n , n + 2 and n + 4 are primes .

Since n is prime and n > 3 we have

n mod 3 = 1 or n mod 3 = 2 . (Why ?)

Case 1 : If n mod 3 = 1 then n = 3k + 1 and

n + 2 = 3k + 3 , i.e., 3|(n + 2) .

Case 2 : If n mod 3 = 2 then n = 3k + 2 and

n + 4 = 3k + 6 , i.e., 3|(n + 4) .Contradiction !

200

THE FACTORIZATION THEOREM :

∀n ∈ (Z+ − {1}) , ∃!(

m, {pi}mi=1, {ni}m

i=1

)

:

m ∈ Z+,

∀i (i = 1, · · · , m) : pi, ni ∈ Z+ ,

1 < p1 < p2 < · · · < pm ,

∀i (i = 1, · · · , m) : pi is a prime number ,

n = pn11 pn2

2 · · · pnm

m .

EXAMPLE : 252 = 22 32 71 .

201

PROPOSITION : log23 is irrational.

PROOF : By contradiction:

Suppose log23 is rational, i.e., ∃p, q ∈ Z+ , such that

log23 = p/q .

By definition of the log function it follows that

2p/q = 3 ,

from which2p = 3q .

Let n = 2p . Then n ∈ Z+ , with n ≥ 2 .

Then n has two different prime factorizations, namely

n = 2p and n = 3q .

This contradicts the Factorization Theorem. QED !

202

REMARK :

The fact that

2p 6= 3q ,

also follows from the facts that 2p is even and 3q is odd.

203

DEFINITION :

We call n ∈ Z+ a perfect square if

∃k ∈ Z+ : n = k2 .

FACT :

The factorization of a perfect square has only even powers :

Ifk = pn1

1 pn22 · · · pnm

m ,

thenn = k2 = p2n1

1 p2n22 · · · p2nm

m .

204

PROPOSITION :

If n ∈ Z+ is not a perfect square then

√n is irrational.

PROOF :

By contradiction :

Suppose n is not a perfect square, but√

n is rational.

Thus

√n =

p

q,

i.e.,

p2 = n q2 ,

for some p, q ∈ Z+ .

205

p2 = n q2

• The prime factorization of p2 has even powers only.

• The prime factorization of q2 has even powers only.

• The prime factorization of n must have an odd power,

(otherwise n would be a perfect square).

• Thus the factorization of nq2 must have an odd power.

• Thus p2 has two distinct factorizations:

one with even powers and one with at least one odd power.

This contradicts the uniqueness in the Factorization Theorem.

QED !

206

DEFINITION :

k ∈ Z+ is the greatest common divisor of n,m ∈ Z

+ ,

k = gcd(n,m) ,

if

• k|n and k|m ,

• no positive integer greater than k divides both n and m .

207

REMARK :

One can determinegcd( n , m )

from the minimum exponents in the prime factorizations of n and m.

EXAMPLE : If

n = 168 , m = 900 ,

then

n = 23 31 71 , m = 22 32 52 ,

andgcd(168, 900) = 22 31 = 12 .

208

THE EUCLIDEAN THEOREM :

Let n , d ∈ Z+ , and let

r = n mod d , (the remainder)

Then

gcd( n , d ) =

gcd( d , r ) if r > 0 ,

d if r = 0 .

209

EXAMPLE :

gcd(93 , 36) = gcd(36 , 93 mod 36) = gcd(36 , 21)

= gcd(21 , 36 mod 21) = gcd(21 , 15)

= gcd(15 , 21 mod 15) = gcd(15 , 6)

= gcd(6 , 15 mod 6) = gcd(6 , 3)

= 3 .

210

EXAMPLE :

gcd(2008 , 1947) = gcd(1947 , 2008 mod 1947) = gcd(1947 , 61)

= gcd(61 , 1947 mod 61) = gcd(61 , 56)

= gcd(56 , 61 mod 56) = gcd(56 , 5)

= gcd(5 , 56 mod 5) = gcd(5 , 1)

= 1 .

Thus 2008 and 1947 are relatively prime .

211

LEMMA :

Let a, b, c ∈ Z , and d ∈ Z+ .

Then

(1) ( a = b + c , d|a , d|b ) ⇒ d|c ,

(2) ( a = b + c , d|b , d|c ) ⇒ d|a ,

(3) ( a = bc , d|c ) ⇒ d|a .

212

gcd( n , d ) =

{

gcd( d , r ) if r > 0d if r = 0

PROOF OF THE EUCLIDEAN THEOREM :


n = q · d + r ,where

q = n div d and r = n mod d .

Case 1 : r = 0 .

Then clearly d|n .

Also d|d and no greater number than d divides d .

Hence d = gcd(n, d) .

214

gcd( n , d ) =

{

gcd( d , r ) if r > 0d if r = 0

Case 2 : r > 0 :

Let k = gcd(n, d) .

Then k|n and k|d .


n = q · d + r ,

By Lemma (3) k|qd .

By Lemma (1) k|r .

Thus k divides both d and r .

215

gcd( n , d ) =

{

gcd( d , r ) if r > 0d if r = 0

k = gcd(n, d) , n = q · d + r .

Show k is the greatest common divisor of d and r :

By contradiction :

Suppose k1 > k and k1 = gcd(d, r) .

Thus k1|d and k1|rBy Lemma (3) k1|qd .

By Lemma (2) k1|n .

Thus k1 divides both n and d .

Since k1 > k this contradicts that k = gcd(n, d) . QED !

216

REVIEW EXERCISES.

Problem 1.

Prove that a composite number n has a factor k ≤ √n .

Thus to check if a number n is prime one needs only check whether

n mod k = 0 , k = 2, 3, · · · , ⌊√n⌋ .

Problem 2. Use the above fact to check whether 143 is prime.

217

Problem 3. Find all integer solutions of

2x ≡ 7(mod 17) .

Problem 4. Find all integer solutions of

4x ≡ 5(mod 9) .

Problem 5.

Does there exists an integer x that simultaneously satisfies

x ≡ 2(mod 6) and x ≡ 3(mod 9) ?

218

SOLUTION Problem 3 : From the given equation

2x ≡ 7(mod 17)

it follows that2x mod 17 = 7 ,

so2x = 17q + 7.

Thus 17q + 7 must be even, and hence q must be odd, q = 2k + 1 .

It follows that

2x = 17(2k + 1) + 7 = 34k + 24 ,

from whichx = 17k + 12 ,

that is,x = 12(mod 17) .

SOLUTION Problem 4 : From the given equation

4x ≡ 5(mod 9) .

it follows that4x mod 9 = 5 ,

so4x = 9q + 5 .

Thus 9q + 5 must be even and hence q must be odd, q = 2k + 1 .

This gives4x = 9(2k + 1) + 5 = 18k + 14 ,

from which2x = 9k + 7 ,

which must be even. Thus k must be odd, k = 2s + 1 . We get

2x = 9(2s + 1) + 7 = 18s + 16 ,

from whichx = 9s + 8 .

Thusx = 8( mod 9) .

SOLUTION Problem 5 : From

x = 2( mod 6) , and x = 3( mod 9) ,

it follows that

x mod 6 = 2 , and x mod 9 = 3 .

Hencex = 6k + 2 and x = 9s + 3 ,

so that6k + 2 = 9s + 3 ,

from which6k − 9s = 1 .

Thus3(2k − 3s) = 1 ,

which clearly has no solutions.

Problem 6. Let

S = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 } ,

and definef : S −→ S ,

byf(k) = (5k + 3) mod 10 .

Is f invertible?

Problem 7. with S as above, also consider

f(k) = (6k + 3) mod 10 ,

andf(k) = (7k + 3) mod 10 .

219

Problem 8. Let n ≥ 2 ,

Sn = { 0 , 1 , 2 , 3 , · · · , n − 1 } ,

and define

f : Sn −→ Sn ,

by

f(k) = (pk + s) mod n ,

where p is prime, with p > n , and s ∈ Sn .

Prove that f is one-to-one (and hence onto and invertible ).

220


Suppose

f(k1) = f(k2) , where 0 ≤ k1 < n and 0 ≤ k2 < n .Thus

(pk1 + s) mod n = (pk2 + s) mod n ,

from whichn | [(pk1 + s) − (pk2 + s)] .

Hence

n | [p(k1 − k2)] , where − (n − 1) ≤ k1 − k2 < n − 1 .

Between −(n − 1) and (n − 1) only 0 is divisible by n.

Also, no factor of n divides divides p, because p is prime, with p > n .

Hence k1 − k2 = 0, i.e., k1 = k2 .

Thus f is one-to-one.

THE PRINCIPLE OF INDUCTION

LetS = { s1 , s2 , s3 , · · · }

be a countably infinite set .

Suppose P is a predicate,

P : S −→ { T , F } ,

such that :

(i) P (s1) = T ,

(ii) for all n : P (sn) = T ⇒ P (sn+1) = T .

ThenP (s) = T , for all s ∈ S .

221

EXAMPLE :

n∑

k=1

k =n(n + 1)

2, ∀n ∈ Z

+ .

Here S = Z+ and

P (n) = T ifn

∑

k=1

k =n(n + 1)

2,

and

P (n) = F if

n∑

k=1

k 6= n(n + 1)

2.

We must show that P (n) = T for all n .

222

PROOF :

(i) “By inspection” the formula holds if n = 1 , i.e., P (1) = T .

(ii) Suppose P (n) = T for some arbitrary n ∈ Z+ , i.e.,

n∑

k=1

k =n(n + 1)

2, ( “the inductive assumption′′ ) .

We must show that P (n + 1) = T , i.e.,

n+1∑

k=1

k =(n + 1)

(

(n + 1) + 1)

2, ( “the inductive step′′ ) .

This is done as follows:

n+1∑

k=1

k =(

n∑

k=1

k)

+ (n + 1) =n(n + 1)

2+ (n + 1)

=n(n + 1) + 2(n + 1)

2=

(n + 1)(

(n + 1) + 1)

2. QED !

223

EXAMPLE :n

∑

k=1

k2 =n(n + 1)(2n + 1)

6, ∀n ∈ Z

+ .

PROOF :

(i) Again the formula is valid if n = 1 .

(ii) Supposen

∑

k=1

k2 =n(n + 1)(2n + 1)

6,

for some arbitrary n ∈ Z+ .

We must show that

n+1∑

k=1

k2 =(n + 1)

(

(n + 1) + 1) (

2(n + 1) + 1)

6.

224

∑nk=1 k2 = n(n+1)(2n+1)

6⇒ ∑n+1

k=1 k2 = (n+1) ((n+1)+1) (2(n+1)+1)6

To do this :

n+1∑

k=1

k2 =(

n∑

k=1

k2)

+ (n + 1)2

= n(n + 1)(2n + 1)/6 + (n + 1)2

= (n + 1)(

n(2n + 1) + 6(n + 1))

/6

= (n + 1) (2n2 + 7n + 6)/6

= (n + 1) (n + 2) (2n + 3)/6

= (n + 1)(

(n + 1) + 1) (

2(n + 1) + 1)

/6 . QED !

225

EXAMPLE :

(n3 − n) mod 3 = 0 , ∀n ∈ Z+ .

( Can you prove this by a direct proof ?)

PROOF ( by induction ) :

(i) By inspection P (1) = T .

(ii) Suppose P (n) = T , i.e.,

(n3 − n) mod 3 = 0 , for some arbitrary n ∈ Z+ .

We must show that P (n + 1) = T , i.e.,(

(n + 1)3 − (n + 1))

mod 3 = 0 .

226

(n3 − n) mod 3 = 0 ⇒ ( (n + 1)3 − (n + 1) ) mod 3 = 0

To do this :

(

(n + 1)3 − (n + 1))

mod 3 = (n3 + 3n2 + 3n − n) mod 3

=(

3(n2 + n) + n3 − n)

mod 3

= (n3 − n) mod 3 = 0 . QED !

227

EXAMPLE :

Let P (n) denote the statement

“A set Sn of n elements has 2n subsets”.

CLAIM : P (n) = T for all n ≥ 0 .

PROOF :

(i) P (0) = T because the empty set has one subset, namely itself.

(ii) Suppose that P (n) = T for some arbitrary n , ( n ≥ 0 ) ,

i.e., Sn has 2n subsets.

We must show that P (n+1) = T , i.e., Sn+1 has 2n+1 subsets.

228

To do this write

Sn+1 = { s1 , s2 , · · · , sn , sn+1 } = Sn ∪ {sn+1} .

Now count the subsets of Sn+1 :

(a) By inductive hypothesis Sn has 2n subsets.

These are also subsets of Sn+1 .

(b) All other subsets of Sn+1 have the form

T ∪ {sn+1} ,

where T is any subset of Sn .

Thus there are 2n such additional subsets.

The total number of subsets of Sn+1 is therefore

2n + 2n = 2n+1 . QED !

229

EXAMPLE :


3n < n!

CLAIM :

P (n) = T for all integers n with n > 6 .

REMARK : P (n) is False for n ≤ 6 . ( Check! )

230

PROOF :

(i) P (7) = T , because

37 = 2187 < 5040 = 7!

(ii) Assume P (n) = T for some arbitrary n , (n ≥ 7) ,

i.e.,3n < n! (n ≥ 7) .

We must show that P (n + 1) = T , i.e.,

3n+1 < (n + 1)!

231

3n < n! ⇒ 3n+1 < (n + 1)!

To do this :

3n+1 = 3 · 3n

< 3 · n! (by inductive assumption)

< (n + 1) n! (since n ≥ 7)

= (n + 1)!

232

EXERCISE : For which nonnegative integers is

n2 ≤ n! ?

Prove your answer by induction.

EXERCISE : For which positive integers n is

n2 ≤ 2n ?

Prove your answer by induction.

233

SOLUTION : n2 ≤ n! , for all n ≥ 4 .

PROOF : If n = 4 then n2 = 16 and n! = 24 so the inequality holds.

Inductively assume that for arbitrary n , (n ≥ 4) , we have n2 ≤ n! .

We must show that (n + 1)2 ≤ (n + 1)! .

To do this :

(n + 1)2 = n2 + 2n + 1 ≤ n! + n! = 2n! < (n + 1)n! = (n + 1)! ,

where we used the inductive assumption and the fact that

2n + 1 ≤ n! for n ≥ 4 .

The latter is easily seen as follows :

2n + 1 < 2n + n = 3n ≤ n! for n ≥ 4 .

QED !

SOLUTION : n2 ≤ 2n , for all n ≥ 4 .

PROOF : If n = 4 then n2 = 16 and 2n = 16 so the inequality holds.

Inductively assume that for arbitrary n , (n ≥ 4) , we have n2 ≤ 2n .

We must show that (n + 1)2 ≤ 2n+1 .

To do this :

(n + 1)2 = n2 + 2n + 1 ≤ n2 + n2 = 2n2 ≤ 2 · 2n = 2n+1 ,

where we used the inductive assumption and the fact that

2n + 1 ≤ n2 for n ≥ 4 ,i.e., that

f(n) ≡ n2 − 2n − 1 ≥ 0 for n ≥ 4 .

The latter is easily seen using Calculus :

f(4) = 7 and f ′(n) = 2(n − 1) > 0 for n > 1 .

QED !

EXAMPLE :

Let

Hm ≡m

∑

k=1

1

k. (”Harmonic numbers.”)


H2n ≥ 1 +n

2.

CLAIM : P (n) = T for all n ∈ Z+ .

PROOF :

(i) It is clear that P (1) = T , because

H21 =

2∑

k=1

1

k= 1 +

1

2.

234

(ii) Assume that

P (n) = T for some arbitrary n ∈ Z+ ,

i.e.,

H2n ≥ 1 +n

2.

We must show that

P (n + 1) = T ,

i.e.,

H2n+1 ≥ 1 +n + 1

2.

235

To do this :

H2n+1 =2n+1∑

k=1

1

k=

2n

∑

k=1

1

k+

2n+1∑

k=2n+1

1

k

=

2n

∑

k=1

1

k+

2n+2n

∑

k=2n+1

1

k

≥ (1 +n

2) + 2n 1

2n + 2n

= (1 +n

2) +

1

2

= 1 +n + 1

2. QED !

236

REMARK :

It follows that

∞∑

k=1

1

kdiverges ,

i.e.,

n∑

k=1

1

k−→ ∞ as n −→ ∞ .

237

EXAMPLE : (The Binomial Formula.)

For n ≥ 0 , a, b nonzero,

(a + b)n =

n∑

k=0

(

nk

)

an−k bk,

where

(

nk

)

≡ n!

k! (n − k)!.

REMARK : Thus we can write

(a+ b)n = an +

(

n1

)

an−1b+

(

n2

)

an−2b2 + · · ·+(

nn − 1

)

abn−1 + bn .

238

PROOF :

The formula holds if n = 0 . ( Check! )

Assume that for some arbitrary n, ( n ≥ 0 )

(a + b)n =

n∑

k=0

(

nk

)

an−k bk .

We must show that the formula is also valid for n + 1 , i.e., that

(a + b)n+1 =n+1∑

k=0

(

n + 1k

)

an+1−k bk .

239

This can be done as follows :

(a + b)n+1 = (a + b)(a + b)n = (a + b)n

∑

k=0

(

nk

)

an−kbk

= an+1 +

n∑

k=1

(

nk

)

an−k+1bk +

n−1∑

k=0

(

nk

)

an−kbk+1 + bn+1

= an+1 +

n∑

k=1

(

nk

)

an−k+1bk +

n∑

k=1

(

nk − 1

)

an−k+1bk + bn+1

= an+1 +n

∑

k=1

{

(

nk

)

+

(

nk − 1

)

}

an−k+1bk + bn+1

= an+1 +n

∑

k=1

(

n + 1k

)

an−k+1bk + bn+1

=

n+1∑

k=0

(

n + 1k

)

an+1−kbk . QED !

240

REMARK :

In the proof we used the fact that

(

nk

)

+

(

nk − 1

)

=n!

k! (n − k)!+

n!

(k − 1)! (n − k + 1)!

=n! (n − k + 1) + n! k

k! (n − k + 1)!

=(n + 1)!

k! (n − k + 1)!

=

(

n + 1k

)

.

241

REMARK :

One can order the binomial coefficients in Pascal’s triangle as follows :

11 1

1 2 11 3 3 1

1 4 6 4 11 5 10 10 5 1

1 6 15 20 15 6 11 7 21 35 35 21 7 1

1 8 28 56 70 56 28 8 1. . . . . . . . . .

242

Observe that every entry can be obtained by summing the closest entriesin the preceding row.

This is so because the (n + 1)st and (n + 2)nd rows look like :

1 · · · · · · · · ·(

nk − 1

) (

nk

)

· · · · · · · · · 1

1 · · · · · · · · ·(

n + 1k

)

· · · · · · · · · 1

and we have shown above that

(

nk

)

+

(

nk − 1

)

=

(

n + 1k

)

.

243

REMARKS :

• The advantage of a proof by induction is that it is systematic.

• A disadvantage is that the result (e.g., a formula) must be known inadvance from a heuristic argument or by trial and error.

• In contrast, a constructive proof actually derives the result.

244

EXAMPLE : For x ∈ R , x 6= 0, 1 :n

∑

k=0

xk =1 − xn+1

1 − x, ∀n ≥ 0 , ( Geometric sum ) .

PROOF ( a constructive proof : already done earlier ) :

LetSn =

n∑

k=0

xk .

Then

Sn = 1 + x + x2 + · · · + xn−1 + xn ,

x · Sn = x + x2 + · · · + xn−1 + xn + xn+1 ,

so thatSn − x · Sn = (1 − x) · Sn = 1 − xn+1 ,

from which the formula follows. QED !

245

ALTERNATE PROOF ( by induction ) :

(i) “By inspection” we find that the formula holds if n = 0 .

(ii) Suppose that Sn = 1−xn+1

1−x, for some arbitary n, ( n ≥ 0 ) .

Show Sn+1 = 1−x(n+1)+1

1−x:

Sn+1 = Sn + xn+1 =1 − xn+1

1 − x+ xn+1

=(1 − xn+1) + xn+1(1 − x)

1 − x

=1 − xn+1 + xn+1 − xn+2

1 − x

=1 − x(n+1)+1

1 − x. QED !

246

EXERCISE :

Use mathematical induction to prove that

21 | (4n+1 + 52n−1) ,

whenever n is a positive integer.

EXERCISE :

The Fibonacci numbers are defined as: f1 = 1 , f2 = 1 , and

fn = fn−1 + fn−2 , for n ≥ 3 .

Use a proof by induction to show that

3 | f4n ,

for all n ≥ 1 .

247

SOLUTION : 21 | (4n+1 + 52n−1) for n ∈ Z+ .

PROOF : This is clearly true if n = 1 , because in that case

4n+1 + 52n−1 = 16 + 5 = 21 .

Inductively assume that for arbitrary n ∈ Z+ we have 21 | (4n+1 + 52n−1) .

We must show that 21 | (4n+2 + 52n+1) .

To do this :

4n+2 + 52n+1 = 4(4n+1 + 52n−1) − 4 · 52n−1 + 52n+1

= 4(4n+1 + 52n−1) − 4 · 52n−1 + 25 · 52n−1

= 4(4n+1 + 52n−1) + 21 · 52n−1 .

where, in the last line, the second term is clearly divisible by 21 ,

while the first term is divisible by 21 by inductive assumption.

QED !

SOLUTION : The Fibonacci numbers defined as

f1 = 1 , f2 = 1 , and fn = fn−1 + fn−2 , for n ≥ 3 ,satisfy

3 | f4n , for all n ∈ Z+ .

PROOF : This is clearly true if n = 1 , because in that case

f4n = f4 = f3 + f2 = (f2 + f1) + f2 = (1 + 1) + 1 = 3 .

Inductively assume that for arbitrary n ∈ Z+ we have 3 | f4n .

We must show that 3 | f4(n+1) :

f4(n+1) = f4n+4 = f4n+3 + f4n+2

= (f4n+2 + f4n+1) + (f4n+1 + f4n)= f4n+2 + 2f4n+1 + f4n

= (f4n+1 + f4n) + 2f4n+1 + f4n

= 3f4n+1 + 2f4n ,

where both terms in the final line are divisible by 3 . QED !

Variations on the Principle of Induction.

Let S = { s1, s2, s3, · · · } be a countably infinite set and P a predicate :

P : S −→ { T , F } .

VARIATION 1 : ( as used so far · · · )(

P (s1) ∧[

∀n ≥ 1 : P (sn) ⇒ P (sn+1)])

⇒ ∀n : P (sn) .

VARIATION 2 :(

P (s1) ∧ P (s2) ∧[

∀n ≥ 2 : P (sn−1) ∧ P (sn) ⇒ P (sn+1)])

⇒∀n : P (sn) .

VARIATION · · ·

STRONG INDUCTION :(

P (s1) ∧ ∀n ≥ 1 :[

P (s1) ∧ P (s2) ∧ · · · ∧ P (sn) ⇒ P (sn+1)])

⇒∀n : P (sn) .

248

EXAMPLE : The Fibonacci Numbers.

The Fibonacci numbers are defined recursively as

f1 = 1 ,

f2 = 1 ,

fn = fn−1 + fn−2 , for n ≥ 3 .

249

f1 = 1 f11 = 89 f21 = 10946

f2 = 1 f12 = 144 f22 = 17711

f3 = 2 f13 = 233 f23 = 28657

f4 = 3 f14 = 377 f24 = 46368

f5 = 5 f15 = 610 f25 = 75025

f6 = 8 f16 = 987 f26 = 121393

f7 = 13 f17 = 1597 f27 = 196418

f8 = 21 f18 = 2584 f28 = 317811

f9 = 34 f19 = 4181 f29 = 514229

f10 = 55 f20 = 6765 f30 = 832040

250

PROPERTY :

There is an explicit formula for fn , namely

fn =1√5

[ (1 +√

5

2

)n

−(1 −

√5

2

)n ]

.

251

We can also write

fn =1√5

[ (1 +√

5

2

)n

−(1 −

√5

2

)n ]

=1√5

(1 +√

5

2

)n [

1 −(1 −

√5

1 +√

5

)n ]

≈ 1√5

(1 +√

5

2

)n [

1 −(1 − 2.236

1 + 2.236

)n ]

=1√5

(1 +√

5

2

)n [

1 + (−1)n+1(

0.3819)n ]

≈ 1√5

(1 +√

5

2

)n

.

252

f1 = 1 ≈ 0.72361 f11 = 89 ≈ 88.99775

f2 = 1 ≈ 1.17082 f12 = 144 ≈ 144.00139

f3 = 2 ≈ 1.89443 f13 = 233 ≈ 232.99914

f4 = 3 ≈ 3.06525 f14 = 377 ≈ 377.00053

f5 = 5 ≈ 4.95967 f15 = 610 ≈ 609.99967

f6 = 8 ≈ 8.02492 f16 = 987 ≈ 987.00020

f7 = 13 ≈ 12.98460 f17 = 1597 ≈ 1596.99987

f8 = 21 ≈ 21.00952 f18 = 2584 ≈ 2584.00008

f9 = 34 ≈ 33.99412 f19 = 4181 ≈ 4180.99995

f10 = 55 ≈ 55.00364 f20 = 6765 ≈ 6765.00003

253

fn = 1√5

[ (

1+√

52

)n

−(

1−√

52

)n ]

PROOF (By Induction, using Variation 2) :

The formula is valid when n = 1 :

f1 =1√5

[ 1 +√

5

2− 1 −

√5

2

]

= 1 .

The formula is also valid when n = 2 :

f2 =1√5

[ (1 +√

5

2

)2

−(1 −

√5

2

)2 ]

=1√5

√5 = 1 .

( Check! )

254

Inductively, assume that we have

fn−1 =1√5

[ (1 +√

5

2

)n−1

−(1 −

√5

2

)n−1 ]

,

and

fn =1√5

[ (1 +√

5

2

)n

−(1 −

√5

2

)n ]

.

We must show that

fn+1 =1√5

[ (1 +√

5

2

)n+1

−(1 −

√5

2

)n+1 ]

.

255

Using the inductive hypothesis for n and n − 1 we have

fn+1 = fn−1 + fn (by definition)

=1√5

[ (1 +√

5

2

)n−1

−(1 −

√5

2

)n−1

+(1 +

√5

2

)n

−(1 −

√5

2

)n ]

=1√5

[ (1 +√

5

2

)n−1 (

1 +1 +

√5

2

)

−(1 −

√5

2

)n−1 (

1 +1 −

√5

2

) ]

=1√5

[ (1 +√

5

2

)n−1 (3 +√

5

2

)

−(1 −

√5

2

)n−1 (3 −√

5

2

) ]

=1√5

[ (1 +√

5

2

)n−1 (1 +√

5

2

)2

−(1 −

√5

2

)n−1 (1 −√

5

2

)2 ]

=1√5

[ (1 +√

5

2

)n+1

−(1 −

√5

2

)n+1 ]

. QED !

256

Direct solution of the Fibonacci recurrence relation.

f1 = 1 , f2 = 1 ,

fk = fk−1 + fk−2 , for k ≥ 3 .

Try solutions of the form

fn = c zn ,

This gives

c zn = c zn−1 + c zn−2 , or zn − zn−1 − zn−2 = 0 ,

from which we obtain the characteristic equation

z2 − z − 1 = 0 .

257

z2 − z − 1 = 0

The characteristic equation has solutions (“roots”):

z =1 ±

√1 + 4

2,

that is,

z1 =1 +

√5

2, z2 =

1 −√

5

2.

The general solution of the recurrence relation is then

fn = c1 zn1 + c2 zn

2 .

258

fn = c1 zn1 + c2 zn

2

The constants c1 and c2 are determined by the initial conditions :

f1 = 1 ⇒ c1 z1 + c2 z2 = 1 ,

andf2 = 1 ⇒ c1 z2

1 + c2 z22 = 1 ,

that is,

c11 +

√5

2+ c2

1 −√

5

2= 1 ,

and

c1

(1 +√

5

2

)2

+ c2

(1 −√

5

2

)2

= 1 ,

from which we find

c1 =1√5

and c2 = − 1√5

. ( Check! )

259

fn = c1 zn1 + c2 zn

2

We found that

z1 =1 +

√5

2, z2 =

1 −√

5

2.

and

c1 =1√5

and c2 = − 1√5

. ( Check! )

from which

fn =1√5

(1 +√

5

2

)n

− 1√5

(1 −√

5

2

)n

.

260

REVIEW EXERCISES.

Problem 1.

Prove that the Fibonacci numbers satisfy the following relations:

• ∑nk=1 f2k−1 = f2n , for n ∈ Z

+ .

• fn−1 fn+1 − f2n = (−1)n , for n ∈ Z

+ , (n ≥ 2) .

• −f1 + f2 − f3 + · · · − f2n−1 + f2n = f2n−1 − 1 , for n ∈ Z+ .

261

SOLUTION :∑n

k=1 f2k−1 = f2n for n ∈ Z+ .

PROOF : This is true if n = 1 , because f1 = f2.

Inductively, assume that for arbitrary n we have

n∑

k=1

f2k−1 = f2n .

We must show n+1∑

k=1

f2k−1 = f2n+2 .

To do this :

n+1∑

k=1

f2k−1 =n

∑

k=1

f2k−1 + f2(n+1)−1

= f2n + f2n+1 = f2n+2 ,

using the inductive assumption and definition of Fibonacci numbers.

QED !

SOLUTION : fn−1 fn+1 − f2n = (−1)n for n ≥ 2 .

PROOF : True if n = 2 , since f1f3 − f22 = 2 − 12 = (−1)2 .

Inductively, assume that for arbitrary n , (n ≥ 2) , we have

fn−1 fn+1 − f2n = (−1)n .

We must showfn fn+2 − f2

n+1 = (−1)n+1 .

Note that it suffices to show that

(fn fn+2 − f2n+1) + (fn−1 fn+1 − f2

n) = 0 .

Indeed, the LHS of this expression can be rewritten as

fn(fn+2 − fn) − fn+1(fn+1 − fn−1) = fnfn+1 − fn+1fn = 0 .

QED !

SOLUTION : −f1 + f2 − f3 + · · · − f2n−1 + f2n = f2n−1 − 1 .

PROOF : True if n = 1 : −f1 + f2 = 0 = f1 − 1 .

Inductively, assume that for arbitrary n we have

−f1 + f2 − f3 + · · · − f2n−1 + f2n = f2n−1 − 1 .

We must show

−f1 + f2 − f3 + · · · − f2n−1 + f2n − f2n+1 + f2n+2 = f2n+1 − 1 .

Indeed , using the inductive assumption and Fibonacci definitions,

(−f1 + f2 − f3 + · · · − f2n−1 + f2n) − f2n+1 + f2n+2

= (f2n−1 − 1) − f2n+1 + f2n+2

= (f2n−1 − 1) + f2n

= f2n+1 − 1 QED !

Problem 2. The recurrence relation

xn+1 = c xn (1 − xn) , n = 1, 2, 3, · · · ,

known as the logistic equation, models population growth when there arelimited resources.

Write a small computer program (using real arithmetic) to see whathappens to the sequence xn, n = 1, 2, 3, · · · , with 0 < x1 < 1 , foreach of the following values of c :

(a) 0.5 , (b) 1.5 , (c) 3.2 , (d) 3.5 , (e) 3.9

Problem 3. Find an explicit solution to the recurrence relation

xn+1 = 3 xn − 2 xn−1 , n = 1, 2, 3, · · · ,

with x1 = 1 and x2 = 3 .

262

RELATIONS

A binary relation relates elements of a set to elements of another set.

EXAMPLE :

The operator “≤” relates elements of Z to elements of Z. e.g.,

2 ≤ 5 , and 3 ≤ 3 , but 7 6≤ 2 .

We can also view this relation as a function

≤ : Z × Z −→ {T ,F } .

263

RECALL :

LetA = {a1, a2, · · · , anA

} and B = {b1, b2, · · · , bnB} .

The product set A × B is the set of all ordered pairs from A and B.

More precisely,

A × B ≡ {(a, b) : a ∈ A, b ∈ B} .

264

NOTE :

• The product set A × B has nA · nB elements.

• If A and B are distinct and nonempty then A × B 6= B × A.

EXAMPLE :

IfA = { ! , ? } and B = { • , ◦ , } ,

then

A × B = { (!, •) , (!, ◦) , (!, ) , (?, •) , (?, ◦) , (?, ) } .

265

We can now equivalently define :

DEFINITION :

A binary relation R from A to B is a subset of A × B.

NOTATION :

If R ⊆ A × B, and(a, b) ∈ R ,

then we say that

“a is R-related to b”,

and we also writea R b .

266

EXAMPLE :

Let A = {1, 3} and B = {3, 5, 9}.

Let R denote the relation “divides” from A to B, i.e.,

aRb if and only if a|b .

Then1R3 , 1R5 , 1R9 , 3R3 , and 3R9 .

ThusR = { (1, 3) , (1, 5) , (1, 9) , (3, 3) , (3, 9) } .

267

We can represent R by the following diagram

3

A B1

3

5

9

R

This representation is an example of a directed bipartite graph.

Note that R is not a function, since it is multi-valued.

268

REMARK :

We see that a relation generalizes the notion of a function.

Unlike functions from a set A to a set B :

• A relation does not have to be defined for all a ∈ A .

• A relation does not have to be single-valued.

269

A finite relation from a set A into itself can be represented

by an ordinary directed graph.

EXAMPLE :

Let

A = {1 , 2 , 3 , 4 , 5 , 6} ,

and let R denote the relation “divides” from A to A .

We say that R is a relation “on A” .

This relation has the following directed graph representation :

270

1

32

56

4

The “divides” relation on the set A .

271

We can compose relations as follows :

Let R be a relation from A to B , and S a relation from B to C.

B C

R

S

S

A

Ro

Then S ◦ R is the relation from A to C defined by

a(S ◦ R)c if and only if ∃b ∈ B : aRb ∧ bSc .

272

EXAMPLE :

Let

A = {1, 2, 3}, B = {2, 6}, and C = {1, 9, 15},

and define the relations R and S by

aRb if and only if a|b ,

and

bSc if and only if b + c is prime .

DefineT = S ◦ R .

273

Then from the diagram below we see that

T = { (1, 1) , (1, 9) , (1, 15) , (2, 1) , (2, 9) , (2, 15) , (3, 1) } .

C

R

S

S

A

Ro

1

2

3

2

6

1

9

15

B

= T

274

Let A, B, C, and D be sets, and let R S, and T be relations :

R : A −→ B , S : B −→ C , T : C −→ D .

PROPOSITION :

The composition of relations is associative, i.e.,

(T ◦ S) ◦ R = T ◦ (S ◦ R) .

A B C

R S T

R

T

S

S R

o

o o

oT S

D

275

A B C

R S T

R

T

S

S R

o

o o

oT S

D

(T ◦ S) ◦ R = T ◦ (S ◦ R) .

PROOF : Let a ∈ A and d ∈ D. Then

a(T ◦ S) ◦ Rd ⇐⇒ ∃b ∈ B : (aRb ∧ bT ◦ Sd)

⇐⇒ ∃b ∈ B, ∃c ∈ C : (aRb ∧ bSc ∧ cTd)

⇐⇒ ∃c ∈ C, ∃b ∈ B : (aRb ∧ bSc ∧ cTd)

⇐⇒ ∃c ∈ C : (aS ◦ Rc ∧ cTd) ⇐⇒ aT ◦ (S ◦ R)d .

QED !

276

EXAMPLE : Let the relation R on

A = { 2, 3, 4, 8, 9, 12 } ,

be defined by

(a, b) ∈ R if and only if (a|b ∧ a 6= b) .

Then

R = { (2, 4) , (2, 8) , (2, 12) , (3, 9) , (3, 12) , (4, 8) , (4, 12) } ,

andR2 ≡ R ◦ R = { (2, 8) , (2, 12) } ,

R3 ≡ R2 ◦ R = R ◦ R ◦ R = { } .

277

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

R R R

RR

2

3

12

3

4

9

3

4

8

9

3

4

8

9

8

4

3

2

9

2

8

12

2

12

2

12

278

EXAMPLE :

Let R be the relation on the set of all real numbers defined by

xRy if and only if xy = 1

Then

xR2z ⇐⇒ ∃y : xRy and yRz

⇐⇒ ∃y : xy = 1 and yz = 1

⇐⇒ x = z and x 6= 0 . (Why ?)

279

The last equivalence in detail:

∃y : xy = 1 and yz = 1 if and only if x = z and x 6= 0 .

PROOF : (⇒) Let x and z be real numbers, and assume that

∃y : xy = 1 and yz = 1 .

Then x and z cannot equal zero.

Thus we can write

y =1

xand y =

1

z.

Hence 1/x = 1/z, i.e., x = z.

Thus x = z and x 6= 0 .

280

∃y : xy = 1 and yz = 1 if and only if x = z and x 6= 0 .

(⇐)

Conversely, suppose x and z are real numbers with

x = z and x 6= 0 .

Let y = 1/x.

Then xy = 1 and yz = 1.

QED !

281

Similarly

xR3z ⇐⇒ ∃y : xR2y and yRz

⇐⇒ ∃y : x = y and x 6= 0 and yz = 1

⇐⇒ xz = 1 (Why ?)

Thus

R3 = R ,

R4 = R3 ◦ R = R ◦ R = R2 ,

R5 = R4 ◦ R = R2 ◦ R = R3 = R ,

and so on · · ·

Thus we see that

Rn = R2 if n is even ,and

Rn = R if n is odd .

282

EXAMPLE :

Let R be the relation on the real numbers defined by

xRy if and only if x2 + y2 ≤ 1 .

Then


⇐⇒ ∃y : x2 + y2 ≤ 1 and y2 + z2 ≤ 1

⇐⇒ x2 ≤ 1 and z2 ≤ 1 (Why ?)

⇐⇒ | x | ≤ 1 and | z | ≤ 1 .

283

y

x x

z

R R2

1 −1

1

1

−1

The relations R and R2 as subsets of R2 .

284

Similarly


⇐⇒ ∃y : | x | ≤ 1 and | y | ≤ 1 and y2 + z2 ≤ 1

⇐⇒ ∃y : | x | ≤ 1 and y2 + z2 ≤ 1 (Why ?)

⇐⇒ | x | ≤ 1 and | z | ≤ 1 (Why ?)

Thus R3 = R2.

285

Similarly

R4 = R3 ◦ R = R2 ◦ R = R3 = R2 ,

R5 = R4 ◦ R = R2 ◦ R = R3 = R2 ,

and so on · · ·

Thus we see that

Rn = R2 for all n ≥ 2 .

286

The relation matrix.

A relation between finite sets can be represented by a relation matrix.

(Also known as the transition matrix).

For a relation R from A to B the relation matrix R has entries

Rij =

0 if (ai, bj) 6∈ R ,

1 if (ai, bj) ∈ R .

287

EXAMPLE :

Reconsider the example where

A = {1, 2, 3}, B = {2, 6}, and C = {1, 9, 15} ,

aRb if and only if a|b, and bSc if and only if b + c is prime .

C

R

S

S

A

Ro

1

2

3

2

6

1

9

15

B

= T

288

C

R

S

S

A

Ro

1

2

3

2

6

1

9

15

B

= T

The relation matrices of R and S are

2 6 1 9 15

R =123

1 11 10 1

and S =26

(

1 1 11 0 0

)

.

289

We found that

T = S ◦R = { (1, 1) , (1, 9) , (1, 15) , (2, 1) , (2, 9) , (2, 15) , (3, 1) } .

C

R

S

S

A

Ro

1

2

3

2

6

1

9

15

B

= T

290

C

R

S

S

A

Ro

1

2

3

2

6

1

9

15

B

= T

The relation matrices of R and S were found to be

R =

1 11 10 1

, S =

(

1 1 11 0 0

)

.

The relation matrix of T = S ◦ R is

T =

1 1 11 1 11 0 0

.

291

PROPOSITION :

Let A, B, and C be finite sets.

Let R be a relation from A to B.

Let S be a relation from B to C.

Let T = S ◦ R.

Then the relation matrix of T has the same zero-structure

as the matrix product RS.

292

PROOF :

Tij = 1 ⇐⇒ aiTcj

⇐⇒ aiRbk and bkScj, for some bk ∈ B

⇐⇒ Rik = 1 and Skj = 1 for some k

⇐⇒ ∑nB

ℓ=1 Riℓ Sℓj 6= 0

⇐⇒ [RS]ij 6= 0 . QED !

293

REMARK :

If we use Boolean arithmetic, then T = RS .

EXAMPLE :

The matrix product RS in the preceding example is

1 11 10 1

(

1 1 11 0 0

)

=

2 1 12 1 11 0 0

.

Using Boolean arithmetic the matrix product is

T =

1 1 11 1 11 0 0

.

294

The inverse of a relation.

Let R be a relation from A to B.

Then the inverse relation R−1 is the relation from B to A defined by

b R−1 a if and only if a R b ,

or, in equivalent notation,

(b, a) ∈ R−1 if and only if (a, b) ∈ R .

Thus, unlike functions, relations are always invertible.

295

EXAMPLE :

1

2

3

a

b

A B

HereR = {(1, a), (1, b), (2, a), (3, b)} ,

andR−1 = {(a, 1), (a, 2), (b, 1), (b, 3)} .

The relation matrices are

R =

1 11 00 1

, R−1 =

(

1 1 01 0 1

)

.

296

Note that

R−1 = RT (transpose) .

This holds in general, because if

A = {a1, a2, · · · , anA} , B = {b1, b2, · · · , bnB

} ,

then, by definition of R−1 we have for any ai, bj that

bjR−1ai ⇐⇒ aiRbj .

Hence [R−1]ji = Rij .

297

EXERCISE : Let R be the relation on the set

A = { 1 , 2 , 3 , 4 } ,

defined bya1Ra2 if and only if a1 < a2 .

• Write down R as a subset of A × A .

• Show the relation matrix of R .

• Do the same for R2 , R3 , · · ·

EXERCISE : Do the same for a1Ra2 if and only if a1 ≤ a2 .

EXERCISE : Do the same for a1Ra2 if and only if a1 + a2 = 5 .

EXERCISE : Do the same for the set Z instead of A .

298

DEFINITION : Let R be a relation on A (i.e., from A to A).

• R is called reflexive if

∀a ∈ A : (a, a) ∈ R, i .e., ∀a ∈ A : aRa ,

i.e., if the relation matrix R (for finite A) satisfies

∀i : Rii = 1 .

EXAMPLES :

The “divides” relation on Z+ is reflexive.

The “≤” relation on Z is reflexive.

The “⊆” relation on a power set 2A is reflexive.

The “<” relation on Z is not reflexive.

299

• R is symmetric if

∀a, b ∈ A : aRb → bRa ,

or, equivalently,aRb ⇒ bRa ,

i.e., if the relation matrix (for finite A) is symmetric :

∀i, j : Rij = Rji .

EXAMPLES :

The relation on the real numbers defined by

xRy if and only if x2 + y2 ≤ 1 ,is symmetric.

The “divides” relation on Z+ is not symmetric.

300

• R is antisymmetric if for all a, b ∈ A we have

aRb ∧ bRa ⇒ a = b ,

or equivalently,

a 6= b ⇒ ¬(aRb) ∨ ¬(bRa) ,

i.e., if the relation matrix (for finite A) satisfies

∀i, j with i 6= j : RijRji 6= 1 .

EXAMPLES :

The “≤” relation on Z is antisymmetric.

The “divides” relation on Z+ is antisymmetric.

The “⊆” relation on a power set 2A is antisymmetric.

301

• R is transitive ifaRb ∧ bRc ⇒ aRc .

We’ll show later that R is transitive if and only if

n∑

k=1

Rk = R

in Boolean arithmetic

EXAMPLES :

The “divides” relation on Z+ is transitive.

The “≤” relation on Z is transitive.

The “⊆” relation on a power set 2S is transitive.

The relation aRb ⇐⇒ “a + b is prime” on Z+ is not transitive.

302

EXERCISE : Let A be a set of n elements.

• How many relations are there on A ?

How many relations are there on A that are :

• symmetric ?

• antisymmetric ?

• symmetric and antisymmetric ?

• reflexive ?

• reflexive and symmetric ?

• transitive (∗) ?

(∗) Hint : Search the web for “the number of transitive relations” !

303

SOLUTIONS :

Let A be a set of n elements.

• The number of relations on A : 2n2

The number of relations on A that are :

• symmetric : 212n(n+1)

• antisymmetric : 2n · 3 12(n−1)n

• symmetric and antisymmetric : 2n

• reflexive : 2n(n−1)

• reflexive and symmetric : 212(n−1)n

• An equivalence relation is a relation that is

- reflexive

- symmetric

- transitive.

EXAMPLE :

The following relation on Z is an equivalence relation :

aRb if and only if a mod m = b mod m .

(Here m ≥ 2 is fixed.)

304

• A partial order is a relation that is

- reflexive

- antisymmetric

- transitive.

EXAMPLES :

The “divides” relation on Z+.

- The “≤” relation on Z+.

- The “⊆” relation on the power set 2S .

- The operator “<” on Z is not a partial order:

(It is antisymmetric (!) and transitive, but not reflexive.)

305

• A relation R on a set A is called a total order if

- R is a partial order, and

- ∀a, b ∈ A we have aRb or bRa .

EXAMPLES :

- The partial order “≤” is also a total order on Z+ .

- The partial order m|n on Z+ is not a total order.

(For example 5 6 |7 and 7 6 |5 .)

- The partial order “⊆” on 2S is not a total order.

306

Equivalence classes.

Let A be a set and let R be an equivalence relation on A .

Let a1 ∈ A.

Define[a1] = {a ∈ A : aRa1} ,

that is

[a1] = all elements of A that “are equivalent ” to a1 .

Then [a1] is called the equivalence class generated by a1 .

307

EXAMPLE :

Let R be the relation “congruence modulo 3” on Z+ , i.e.,

aRb if and only if a mod 3 = b mod 3 .

For example

1R1 , 1R4 , 1R7 , 1R10 , · · · ,

2R2 , 2R5 , 2R8 , 2R11 , · · · ,

3R3 , 3R6 , 3R9 , 3R12 , · · · .

308

Thus

[1] = { 1 , 4 , 7 , 10 , 13 , · · · } ,

[2] = { 2 , 5 , 8 , 11 , 14 , · · · } ,

[3] = { 3 , 6 , 9 , 12 , 15 , · · · } .

We see thatZ

+ = [1] ∪ [2] ∪ [3] .

• The relation R has partitioned Z+ into the subsets [1] , [2] , [3] .

• Any member of a subset can represent the subset, e.g.,

[11] = [2] .

309

EXAMPLE :

Consider Z × Z , the set of all ordered pairs of integers.

Define a relation R on Z × Z by

(a1, b1)R(a2, b2) if and only if a1 − b1 = a2 − b2 .

Note that R can be viewed as subset of

(Z × Z) × (Z × Z) .

310

(a1, b1)R(a2, b2) if and only if a1 − b1 = a2 − b2

R is an equivalence relation :

• R is reflexive : (a, b)R(a, b) ,

• R is symmetric : (a1, b1)R(a2, b2) ⇒ (a2, b2)R(a1, b1) ,

• R is transitive:

(a1, b1)R(a2, b2) ∧ (a2, b2)R(a3, b3) ⇒ (a1, b1)R(a3, b3) .

.

311

(a1, b1)R(a2, b2) if and only if a1 − b1 = a2 − b2

The equivalence classes are

Ak = { (a, b) : a − b = k } .

For example,

A1 = [(2, 1)] = {· · · , (−1,−2) , (0,−1) , (1, 0) , (2, 1) , · · ·} .

The sets Ak partition the set Z × Z , namely,

Z × Z = ∪∞k=−∞ Ak .

312

A

A

A

A

A AA0

1

2

−1−2

3

−3b

a

313

DEFINITION :

• The reflexive closure of R is the smallest relation containing R thatis reflexive.

• The symmetric closure of R is the smallest relation containing Rthat is symmetric.

• The transitive closure of R is the smallest relation containing Rthat is transitive.

314

EXAMPLE :

Let A = {1, 2, 3, 4} , and let R be the relation on A defined by

R = { (1, 4) , (2, 1) , (2, 2) , (3, 2) , (4, 1) } .

Then R is not reflexive, not symmetric, and not transitive :

��

��

��

��

��

��

��

��

2

4 3

1

315

The reflexive closure of R is :

��

��

��

��

��

2

4 3

1

316

The symmetric closure of R is :

��

��

��

��

��

��

2

4 3

1

317

To get the transitive closure of R :

��

��

��

��

��

��

2

4 3

1

318


��

��

��

��

��

��

2

4 3

1

319


��

��

��

��

��

��

2

4 3

1

320

The transitive closure of R is :

��

��

��

��

��

��

2

4 3

1

321

PROPERTY : The transitive closure R∗ of a relation R is given by

R∗ = ∪∞k=1 Rk .

PROOF : Later · · ·

PROPERTY : For a finite set of n elements, the relation matrix ofthe transitive closure is

R∗ =

n∑

k=1

Rk (using Boolean arithmetic) .

NOTE : It suffices to sum only the first n powers of the matrix R !

322

EXAMPLE : In the preceding example,

R = { (1, 4) , (2, 1) , (2, 2) , (3, 2) , (4, 1) } ,

��

��

��

��

��

2

4 3

1

we have the relation matrix :

R =

0 0 0 11 1 0 00 1 0 01 0 0 0

.

323

Thus, using Boolean arithmetic,

R2 = R · R =

0 0 0 11 1 0 00 1 0 01 0 0 0

0 0 0 11 1 0 00 1 0 01 0 0 0

=

1 0 0 01 1 0 11 1 0 00 0 0 1

,

R3 = R · R2 =

0 0 0 11 1 0 00 1 0 01 0 0 0

1 0 0 01 1 0 11 1 0 00 0 0 1

=

0 0 0 11 1 0 11 1 0 11 0 0 0

,

R4 = R · R3 =

0 0 0 11 1 0 00 1 0 01 0 0 0

0 0 0 11 1 0 11 1 0 11 0 0 0

=

1 0 0 01 1 0 11 1 0 10 0 0 1

.

324

Therefore

R∗ =4

∑

k=1

Rk = R + R2 + R3 + R4 =

0 0 0 11 1 0 00 1 0 01 0 0 0

+

1 0 0 01 1 0 11 1 0 00 0 0 1

+

0 0 0 11 1 0 11 1 0 11 0 0 0

+

1 0 0 01 1 0 11 1 0 10 0 0 1

=

1 0 0 11 1 0 11 1 0 11 0 0 1

.

325

The graph of R∗ (shown before) is :

��

��

��

��

2

4 3

1

which indeed has the relation matrix

R∗ =4

∑

k=1

Rk =

1 0 0 11 1 0 11 1 0 11 0 0 1

.

326

RECALL : The transitive closure R∗ of a relation R is given by

R∗ = ∪∞k=1 Rk (to be proved later · · · )

EXAMPLE : Consider the relation R on the real numbers

xRy if and only if xy = 1 .

Earlier we found that

xR2y if and only if x = y ∧ x 6= 0 ,and

R = R3 = R5 = · · · ,

R2 = R4 = R6 = · · · .

Thus the transitive closure is

xR∗y if and only if xy = 1 ∨ (x = y ∧ x 6= 0) .

327

EXAMPLE : Again consider the relation R on the real numbers

xRy if and only if x2 + y2 ≤ 1 .

Earlier we found that

xR2y if and only if | x | ≤ 1 ∧ | y | ≤ 1 ,and

Rn = R2 , for n ≥ 2 .

Thus the transitive closure is

xR∗y if and only if x2 + y2 ≤ 1 ∨ ( | x |≤ 1 ∧ | y |≤ 1 ) ,

that is,

xR∗y if and only if | x | ≤ 1 ∧ | y | ≤ 1 . (Why ?)

328

EXERCISE : Let R be the relation on the real numbers given by

xRy if and only if x2 + y2 = 1 .

• Draw R as a subset of the real plane.

• Is R reflexive?

• Is R symmetric?

• Is R antisymmetric?

• Is R transitive?

• What is R2 ? (Be careful!)

• What is R3 ?

• What is the transitive closure of R ?

329


xRy if and only if xy ≤ 1 .


• Is R reflexive?

• Is R symmetric?



• What is R2 ?

• What is R3 ?


330


xRy if and only if x2 ≤ y .


• Is R reflexive?

• Is R symmetric?



• What is R2 ?

• What is R3 ?

• What is Rn ? (Prove your formula for Rn by induction.)


331

xRy if and only if x2 ≤ y

Then


⇐⇒ ∃y : x2 ≤ y and y2 ≤ z

⇐⇒ x4 ≤ z .

Similarly


⇐⇒ ∃y : x4 ≤ y and y2 ≤ z

⇐⇒ x8 ≤ z .

By induction one can prove that

xRnz ⇐⇒ x2n ≤ z .

332

We see that

• The relation R corresponds to the area of the x, y-plane thatlies on or above the curve y = x2 .

• The relation R2 corresponds to the area of the x, y-plane thatlies on or above the curve y = x4 .

• The relation Rn corresponds to the area of the x, y-plane thatlies on or above the curve y = x2n

.

333

The transitive closure is

R∗ = ∪∞k=1 Rk .

(to be proved later · · · )

We find that R∗ is the union of the following two regions:

• The area of the x, y-plane that lies on or above the curve

y = x2 (i.e., the area that corresponds to the relation R ).

• The area inside the rectangle whose corners are located at

(x, y) = (−1, 0) , (1, 0) , (1, 1) , (−1, 1) ,

(excluding the border of this rectangle).

( Check! )

334

Let R be a relation on a set A .

Recursively define

R1 = R , Rn+1 = Rn ◦ R , n = 1, 2, 3, · · · .

Then for all n ∈ Z+ we have:

PROPERTY 1 : Rn ◦ R = R ◦ Rn

PROPERTY 2 : R symmetric ⇒ Rn is symmetric

EXERCISE : Use induction to prove these properties.

335

PROPERTY 1 : Rn ◦ R = R ◦ Rn, for all n ∈ Z+

PROOF : Clearly, the equality holds if n = 1.

Inductive assumption :

Rn ◦ R = R ◦ Rn , for some arbitrary n ∈ Z+ .

We must show thatRn+1 ◦ R = R ◦ Rn+1 .

To do this :

Rn+1 ◦ R = (Rn ◦ R) ◦ R (by definition)

= (R ◦ Rn) ◦ R (by inductive assumption)

= R ◦ (Rn ◦ R) (by associativity)

= R ◦ Rn+1 (by definition). QED !

336

PROPERTY 2 : R symmetric ⇒ Rn is symmetric

PROOF : Clearly True if n = 1 .Inductively assume that Rn is symmetric.

We must show that Rn+1 is symmetric:

aRn+1b ⇐⇒ aRn ◦ Rb (by definition of power)

⇐⇒ ∃p(aRp ∧ pRnb) (by definition of composition)

⇐⇒ ∃p(pRa ∧ bRnp) (since R and Rn are symmetric)

⇐⇒ ∃p(bRnp ∧ pRa) (commutative law of logic)

⇐⇒ bR ◦ Rna (by definition of composition)

⇐⇒ bRn ◦ Ra (by Property 1)

⇐⇒ bRn+1a (by definition of power) . QED !

337

Let R be a relation on a set A and let n ∈ Z+ .

PROPERTY 3 : R transitive ⇒ Rn is transitive

PROOF :

Let R be transitive.

Obviously R1 is transitive.

By induction assume that Rn is transitive for some n ∈ Z+ .

We must show that Rn+1 is transitive, i.e., we must show that

aRn+1b ∧ bRn+1c ⇒ aRn+1c .

338

Given R and Rn are transitive. Show Rn+1 is transitive

· · · continuation of proof · · ·

aRn+1b ∧ bRn+1c

⇒ aRn ◦ Rb ∧ bRn ◦ Rc (power)

⇒ aRn ◦ Rb ∧ bR ◦ Rnc (by Property 1)

⇒ aRp ∧ pRnb ∧ bRnq ∧ qRc (∃p, q: composition)

⇒ aRp ∧ pRnq ∧ qRc (inductive assumption)

⇒ aRn ◦ Rq ∧ qRc (composition)

⇒ aR ◦ Rnq ∧ qRc (by Property 1)

339

Given R and Rn are transitive. Show Rn+1 is transitive.

· · · continuation of proof · · ·

aR ◦ Rnq ∧ qRc

⇒ aRns ∧ sRq ∧ qRc (∃s: composition)

⇒ aRns ∧ sRc (since R is transitive)

⇒ aR ◦ Rnc (composition)

⇒ aRn ◦ Rc (by Property 1)

⇒ aRn+1c (power) QED !

340

Let R and S be relations on a set A .

Recall that we can also think of R as a subset of A × A .

We have:PROPERTY 4 : R ⊆ S ⇒ Rn ⊆ Sn

PROOF : The statement clearly holds when n = 1 .

Inductive step:

Given R ⊆ S and Rn ⊆ Sn . Show Rn+1 ⊆ Sn+1

To do this :

Suppose that (x, z) ∈ Rn+1 .

Then ∃y : (x, y) ∈ R and (y, z) ∈ Rn .

By the assumptions (x, y) ∈ S and (y, z) ∈ Sn .

Hence (x, z) ∈ Sn+1 . QED !

341

Let S be a relation on a set A .

PROPERTY 5 : S is transitive if and only if ∀n ∈ Z+ : Sn ⊆ S

PROOF :

(⇐) (∀n ∈ Z+ : Sn ⊆ S) ⇒ S is transitive

Let (x, y) ∈ S and (y, z) ∈ S .

Then, by definition of composition, (x, z) ∈ S2 .

Since, in particular, S2 ⊆ S it follows that (x, z) ∈ S .

Hence S is transitive.

342

(⇒) S is transitive ⇒ ∀n ∈ Z+ : Sn ⊆ S

By induction :

Clearly Sn ⊆ S if n = 1 .

Suppose that for some n we have Sn ⊆ S .

We must show that Sn+1 ⊆ S .

343

Given (1): S is transitive, and (2): Sn ⊆ S . Show Sn+1 ⊆ S

To do this, suppose that (x, z) ∈ Sn+1 .

We must show that (x, z) ∈ S .

By definition of composition, (x, z) ∈ Sn ◦ S , and hence

∃y : (x, y) ∈ S and (y, z) ∈ Sn .

By inductive hypothesis (2) (y, z) ∈ S .

Thus (x, y) ∈ S and (y, z) ∈ S .

By assumption (1) S is transitive, so that (x, z) ∈ S . QED !

344

THEOREM :

The transitive closure R∗ of a relation R is given by

R∗ = ∪∞k=1 Rk .

PROOF : Let U = ∪∞k=1 Rk .

We must show that

(1) U is transitive.

(2) U is the smallest transitive relation containing R .

If so, then R∗ = U .

345

U = ∪∞k=1 Rk ⋆

(1) We first show that U is transitive :

Suppose (x, y) ∈ U and (y, z) ∈ U .

We must show that (x, z) ∈ U .

From ⋆ it follows that

(x, y) ∈ Rm and (y, z) ∈ Rn , for some m,n ∈ Z+ .

By definition of composition

(x, z) ∈ Rn+m .

Thus, using ⋆ again, it follows that

(x, z) ∈ U .

346

U = ∪∞k=1 Rk ⋆

(2) Show U is the smallest transitive relation containing R :

To do this it suffices to show that :

( S transitive and R ⊆ S ) ⇒ U ⊆ S .

RU

S

347

U = ∪∞k=1 Rk ⋆

R ⊆ S ⇒ Rn ⊆ Sn Property 4

S is transitive ⇐⇒ ∀n ∈ Z+ : Sn ⊆ S Property 5

To do: Given S transitive and R ⊆ S . Show U ⊆ S

Let (x, y) ∈ U . Then, by ⋆ we have

(x, y) ∈ Rn for some n ∈ Z+ .

By Property 4 : (x, y) ∈ Sn .

By Property 5 : (x, y) ∈ S . QED !

348

REVIEW PROBLEMS

and

REVIEW CLICKER QUESTIONS

349

Review Problem 1.

Prove that for every integer n we have

n5 − n ≡ 0 (mod 30)

350

Review Problem 2.

Suppose m and n are relatively prime integers; m ≥ 2 , n ≥ 2 .

Prove that logmn is an irrational number.

351

Review Problem 3. If A and B are sets, and if

f : A −→ B ,

then for any subset S of B we define the pre-image of S as

f−1(S) ≡ {a ∈ A : f(a) ∈ S} .

NOTE : f−1(S) is defined even if f does not have an inverse!

Let S and T be subsets of B .

Prove that

f−1(S ∩ T ) = f−1(S) ∩ f−1(T )

352

Review Problem 4.

Prove that if a , b , and c are integers such that

m ≥ 2 and a ≡ b(mod m)

then

gcd(a,m) = gcd(b,m) .

353

Review Problem 5.

Use mathematical induction to prove that

21 | (4n+1 + 52n−1) ,

whenever n is a positive integer.

354

Review Problem 6.

The Fibonacci numbers are defined as: f1 = 1 , f2 = 1 , and

fn = fn−1 + fn−2 , for n ≥ 3 .

Use a proof by induction to show that

fn−1 fn+1 − f2n = (−1)n

for all n ≥ 2 .

355

Review Problem 7.

Let A and B be non-empty sets.

Let f be a 1 − 1 function from A to B .

Suppose S is an partial order on B .

Define a relation R on A as follows:

∀a1, a2 ∈ R : a1Ra2 ⇐⇒ f(a1)Sf(a2) .

Prove that R is an partial order on A .

356

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